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CBSE Class 10 Lakhmir Singh Physics
Physics

a) Consider the circuit in the figure. How much energy is absorbed by electrons from the initial state of no current to the state of drift velocity? b) Electrons give up energy at the rate of RI2 per second to the thermal energy. What time scale would one associate with energy in problem a) n = no of electron/volume = 1029/m3, length of circuit = 10 cm, cross-section = A = 1mm2

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a) Current is given as I = V/R from the Ohm’s law Therefore, I = 1A But, I = ne Avd vd = I/neA When the values for the above parameters are substituted, vd = 1/1.6 × 10-4 m/s The KE = (KE of one...

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 In an experiment with a potentiometer, VB = 10V. R is adjusted to be 50 Ω. A student wanting to measure voltage E1 of a battery finds no null point possible. He then diminishes R to 10 Ω and is able to locate the null point on the last segment of the potentiometer. Find the resistance of the potentiometer wire and potential drop per unit length across the wire in the second case.

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Equivalent resistance of the potentiometer = 50 Ohm + R’ Equivalent voltage across the potentiometer = 10 V Current through the main circuit I = 10/(50 Ohms +R’) Potential difference across wire of...

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A room has AC run for 5 hours a day at a voltage of 220V. The wiring of the room consists of Cu of 1 mm radius and a length of 10 m. Power consumption per day is 10 commercial units. What fraction of it goes in the joule heating in wires? What would happen if the wiring is made of aluminium of the same dimensions?

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Power consumption in a day = 10 units Power consumption per hour = 2 units Power consumption = 2 units = 2 kW = 2000 J/s Power consumption in resistors, P = VI Which gives I = 9A We know that...

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Two cells of voltage 10V and 2V and internal resistances 10Ω and 5Ω respectively are connected in parallel with the positive end of the 10V battery connected to the negative pole of 2V battery. Find the effective voltage and effective resistance of the combination.

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Kirchhoff’s law is applied at c, I1 = I + I2 Kirchhoff’s law is applied at efbae, 10 – IR – 10I2 = 0 10 = IR + 10I1 Kirchhoff’s law is applied at cbadc, -2-IR+5I2 = 0 2 = 5I2- RI I2 = I1 – I...

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Two conductors are made of the same material and have the same length. Conductor A is a solid wire of diameter 1 mm. Conductor B is a hollow tube of outer diameter 2 mm and inner diameter 1 mm. Find the ratio of resistance RA to RB.

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Resistance of wire R = ρ l/A Where A is the cross-sectional area of the conductor L is the length of the conductor ρ is the specific resistance RA = ρl/π(10-3 × 0.5)2 RB = ρl/ π[10-3)2 × (0.5 ×...

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The circuit in the figure shows two cells connected in opposition to each other. Cell E1 is of emf 6V and internal resistance 2Ω; the cell E2 is of emf 4V and internal resistance 8 Ω. Find the potential difference between the points A and B.

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Applying Ohm’s law, equivalent emf of the two cells = 6 – 4 = 2V Equivalent resistance = 2 + 8 = 10 Ω Electric current, I = 6-4/2+8 = 0.2A When the loop is considered in the anti-clockwise...

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Two cells of same emf E but internal resistance r1 and r2 are connected in series to an external resistor R. What should be the value of R so that the potential difference across the terminals of the first cell becomes zero.

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Effective emf of two cells = E + E = 2E Effective resistance = R + r1 + r2 Electric current is given as I = 2E/R+r1+r2 Potential difference is given as V1 – E – Ir1 = 0 Which f=gives R = r1 –...

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Let there be n resistors R1……..Rn with Rmax = max(R1……Rn) and Rmin = min(R1…….Rn). Show that when they are connected in parallel, the resultant resistance Rp < Rmin and when they are connected in series, the resultant resistance Rs > Rmax. Interpret the result physically.

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  The current is represented as I = E/R+nR when the resistors are connected in series. Current is expressed as 10I = E/(R+R/n) when the resistors are connected in parallel....

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While doing an experiment with potentiometer it was found that the deflection is one-sided and i) the deflection decreased while moving from one end A of the wire to the end B; ii) the deflection increased, while the jockey was moved towards the end B. i) Which terminal +ve or –ve of the cell E, is connected at X in case
i) and how is E1 related to E?
ii) Which terminal of the cell E1 is connected at X in case ii)?

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The positive terminal of cell E1 is linked to E, and E is connected to X.  Furthermore, E1 > E ii) cell E1's negative terminal is linked to X.

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The relaxation time τ is nearly independent of applied E field whereas it changes significantly with temperature T. First fact is responsible for Ohm’s law whereas the second fact leads to a variation of ρ with temperature. Elaborate why?

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Relaxation time is the time interval between two successive collisions of the electrons.It is defined asτ = mean free path/rms velocity of electrons usually, the drift velocity of the electrons is...

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In a meter bridge, the point D is a neutral point.
a) the meter bridge can have no other neutral point for this set of resistances
b) when the jockey contacts a point on meter wire left of D, current flows to B from the wire
c) when the jockey contacts a point on a meter wire to the right of D, current flows from B to the wire through the galvanometer
d) when R is increased, the neutral point shifts to left

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The correct answer is a) the meter bridge can have no other neutral point for this set of resistances c) when the jockey contacts a point on a meter wire to the right of D, current flows from B to...

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The measurement of an unknown resistance R is to be carried out using Wheatstone bridge. Two students perform an experiment in two ways. The first student take R2 = 10Ω and R1 = 5Ω. The other student takes R2 = 1000 Ω and R1 = 500 Ω. In the standard arm, both take R3 = 5 Ω. Both find R = R2/R1 R3 = 10 Ω within errors.
a) the errors of measurement of the two students are the same
b) errors of measurement do depend on the accuracy with which R2 and R1 can be measured
c) if the student uses large values of R2 and R1, the currents through the arms will be feeble. This will make determination of null point accurately more difficult
d) Wheatstone bridge is a very accurate instrument and has no errors of measurement

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The correct answer is b) errors of measurement do depend on the accuracy with which R2 and R1 can be measured c) if the student uses large values of R2 and R1, the currents through the arms will be...

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Temperature dependence of resistivity ρ(T) of semiconductors, insulators, and metals is significantly based on the following factors:
a) number of charge carriers can change with temperature T
b) time interval between two successive collisions can depend on T
c) length of material can be a function of T
d) mass of carriers is a function of T

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solution:The correct answer is a) number of charge carriers can change with temperature T b) time interval between two successive collisions can depend on T

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Consider a simple circuit in the figure.stands for a variable resistance R’.
R’ can vary from R0 to infinity. r is internal resistance of the battery,
a) potential drop across AB is nearly constant as R’ is varied
b) current through R’ is nearly a constant as R’ is varied
c) current I depends sensitively on R’
d) I ≥V/r+R always

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solution: The correct answer is a) potential drop across AB is nearly constant as R’ is varied d) I ≥V/r+R always  

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Kirchhoff’s junction rule is a reflection of
a) conservation of current density vector
b) conservation of charge
c) the fact that the momentum with which a charged particle approaches a junction is unchanged as the charged particle leaves the junction
d) the fact that there is no accumulation of charges at a junction

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solution: The correct answer is b) conservation of charge d) the fact that there is no accumulation of charges at a junction

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A metal rod of length 10 cm and a rectangular cross-section of 1 cm × 1/2 cm is connected to battery across opposite faces. The resistance will be
a) maximum when the battery is connected across 1 cm × 1/2 cm faces
b) maximum when the battery is connected across 10 cm × 1 cm faces
c) maximum when the battery is connected across 10 cm × 1/2 cm faces
d) same irrespective of the three faces

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solution:The correct solution is a) maximum when the battery is connected across 1 cm × 1/2 cm faces

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Two cells of emf’s approximately 5V and 10V are to be accurately compared using a potentiometer of length 400 cm.
a) the battery that runs the potentiometer should have voltage of 8V
b) the battery of potentiometer can have a voltage of 15V and R adjusted so that the potential drop across the wire slightly exceeds 10V
c) the first portion of 50 cm of wire itself should have a potential drop of 10V
d) potentiometer is usually used for comparing resistances and not voltages

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Solution: The correct solution is b) the potentiometer's battery can be set to 15V and R adjusted so that the potential drop across the wire is a little higher than 10V.

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A resistance R is to be measured using a meter bridge. Student chooses the standard resistance S to be 100Ω. He finds the null point at l1 = 2.9 cm . He is told to attempt to improve the accuracy. Which of the following is a useful way?
a) he should measure l1 more accurately
b) he should change S to 1000 Ω and repeat the experiment
c) he should change S to 3 Ω and repeat the experiment
d) he should give up hope of a more accurate measurement with a meter bridge

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solution:The correct answer is c) he should change S to 3 Ω and repeat the experiment

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Two batteries of emf ε1 and ε2 and internal resistances r1 and r2 respectively are connected in parallel as shown in the figure.a) the equivalent emf εeq of the two cells is between ε1 and ε2 that is ε1 < εeq < ε2
b) the equivalent emf εeq is smaller than ε1
c) the εeq is given by εeq = ε1 + ε2 always
d) εeq is independent of internal resistances r1 and r2

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      solution: The correct answer is a) the equivalent emf εeq of the two cells is between ε1 and ε2 that is ε1 < εeq < ε2

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Consider a current-carrying wire in the shape of a circle. Note that as the current progresses along the wire, the direction of j changes in an exact manner, while the current I remain unaffected. The agent that is essentially responsible for is
a) source of emf
b) electric field produced by charges accumulated on the surface of wire
c) the charges just behind a given segment of wire which push them just the right way by repulsion
d) the charges ahead

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solution: The correct answer is b) electric field produced by charges accumulated on the surface of wire

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A fix number of spherical drops of a liquid of radius ‘ $r$ ‘ coalesce to form a large drop of radius ‘ $\mathrm{R}^{\prime}$ and volume ‘ $\mathrm{V}^{\prime}$. If ‘ $\mathrm{T}^{\prime}$ is the surface tension then energy
A) is niether released nor absorbed
B)$3 \mathrm{VT}\left(\frac{1}{\mathrm{r}}-\frac{1}{\mathrm{R}}\right)$ is released.
C) $4 \mathrm{VT}\left(\frac{1}{\mathrm{r}}-\frac{1}{\mathrm{R}}\right)$ is released.
D) 3 VT $\left(\frac{1}{r}-\frac{1}{R}\right)$ is absorbed.

Correct option is B. $\begin{array}{l} \Delta \mathrm{U}=(\mathrm{T})(\Delta \mathrm{A}) \\ \mathrm{A}(\text { initial })=\left(4 \pi \mathrm{r}^{2}\right) \mathrm{n} \\ \mathrm{A}(\text { final...

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When a beam of unpolarised monochromatic light is incident on a plane glass plate at a polarising angle, then which one of the following statements is correct?
A) Reflected and refracted rays are completely polarised with their planes of polarisation perpendicular to each other.
B) Reflected light is partially polarised but refracted light is plane polarised.
C) Reflected and refracted rays are completely polarised with their planes of polarisation parallel to each other.
D) Reflected light is plane polarised light but transmitted light is partially polarised.

Correct option is D.

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A smooth sphere of mass ‘M’ moving with velocity ‘u’ directly collides elastically with another sphere of mass ‘ $\mathrm{m}$ ‘ at rest. After collision, their final velocities are V’ and $V$ respectively. The value of $V$ is given by
A)$\frac{2 \mathrm{u}}{1+\frac{\mathrm{M}}{\mathrm{m}}}$
B) $\frac{2 \mathrm{um}}{\mathrm{M}}$
C)$\frac{2 \mathrm{u}}{1+\frac{\mathrm{m}}{\mathrm{M}}}$
D)$\frac{2 \mathrm{u} \mathrm{M}}{\mathrm{m}}$

Correct option is C. Using Newton's law of collision, $\begin{array}{l} \frac{\mathrm{V}-\mathrm{v}}{\mathrm{u}-0}=-\mathrm{e} \\ \frac{\mathrm{V}-\mathrm{v}}{\mathrm{u}-0}=-1 \quad \quad...

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A body performs S.H.M. due to force ${ }^{\prime} \mathrm{F}_{1}^{\prime}$, with time period $0.8 \mathrm{~s}$. If force is changed to ${ }^{\prime} \mathrm{F}_{2}{ }^{\prime}$, it executes S.H.M. with time period $0.6 \mathrm{~s}$. Now both the forces act simultaneously in the same direction on the same body. New periodic time is
A) $0.48 \mathrm{~s}$
B)$0.24 \mathrm{~s}$
C)$0.12 \mathrm{~s}$
D)$0.36 \mathrm{~s}$

Correct option is A. $\ln \mathrm{SHM}, F=\frac{4 \pi^{2} m}{T^{2}} y$ or $T^{2} \propto \frac{1}{F}$ or $T \propto \frac{1}{\sqrt{F}}: \operatorname{So} 0.8 \propto \frac{1}{\sqrt{F_{1}}}$ and $0.6...

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An ideal gas occupies a volume ‘ $\mathrm{V}$ ‘ at a pressure ‘ $\mathrm{P}$ ‘ and absolute temperature $\mathrm{T}$. The mass of each molecule is ‘ $\mathrm{m}$ ‘. If ‘ $\mathrm{K}_{\mathrm{B}}{ }^{\prime}$ is the Boltzmann’s constant, then the density of gas is given by expression
A)$\frac{\mathrm{K}_{\mathrm{B}} \cdot \mathrm{T}}{\mathrm{P} \cdot \mathrm{m}}$
B)$\frac{3 \mathrm{~K}_{\mathrm{B}} \cdot \mathrm{T}}{2 \mathrm{P} \cdot \mathrm{m}}$
C)$\frac{\mathrm{P} \cdot \mathrm{m}}{2 \mathrm{~K}_{\mathrm{B}} \cdot \mathrm{T}}$
D)$\frac{\mathrm{P} \cdot \mathrm{m}}{\mathrm{K}_{\mathrm{B}} \cdot \mathrm{T}}$

Correct option is D. The ideal gas law is given by $\mathrm{PV}=\mathrm{nRT}$, where $\mathrm{n}$ is the number of moles of a gas. $\mathrm{n}=$ mass of the gas $/$ molecular weight of the gas....

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The depth of an ocean is $2000 \mathrm{~m}$. The compressibility of water is $45 \times 10^{-11} \mathrm{~m}^{2} / \mathrm{N}$ and density of water is $10^{3} \mathrm{~kg} / \mathrm{m}^{3}$. At the bottom of the ocean, the fractional compression of water will be $\left(\mathrm{g}=10 \mathrm{~m} / \mathrm{s}^{2}\right)$
A)$6 \times 10^{-3}$
B)$10^{-3}$
C)$9 \times 10^{-3}$
D)$3 \times 10^{-3}$

Correct option is C. compressibility is given as $\kappa=\frac{\frac{\Delta V}{V}}{\Delta P}$ $\Delta \mathrm{V}=\kappa \times \Delta \mathrm{P} \times \mathrm{V}$ Substituting values $\Delta...

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For an ideal gas, if the ratio of Molar specific heats $\gamma=1 \cdot 4$, then the specific heat at constant pressure $C_{p}$, specific heat at constant volume $C_{v}$ and corresponding molecule are respectively
A)$\frac{5}{2} \mathrm{R}, \frac{3}{2} \mathrm{R}$, monoatomic.
B)$\frac{9}{2} R, \frac{7}{2} R$, polyatomic.
C)$\frac{7}{2} \mathrm{R}, \frac{5}{2} \mathrm{R}$, non-rigid diatomic.
D) $\frac{7}{2} \mathrm{R}, \frac{5}{2} \mathrm{R}$, rigid diatomic.

Correct option is D. Given, $\gamma=1.4=\frac{7}{5}$ We know, $C_p-C_v=R$ $\gamma=\frac{7}{5}=\frac{C_p}{C_v}$ $\therefore C_p=\frac{7}{5}C_v$ $\frac{7}{5}C_v-C_v=R$ $\frac{2}{5}C_v=R$...

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A uniform wire has length ‘ $\mathrm{L}^{\prime}$ and weight ‘ $\mathrm{W}^{\prime}$. One end of the wire is attached rigidly to a point in the roof and weight ‘ $\mathrm{W}_{1}{ }^{\prime}$ is suspended from its lower end. If ‘ $\mathrm{A}^{\prime}$ is the cross-sectional area of the wire then the stress in the wire at a height $\frac{3 \mathrm{~L}}{4}$ from its lower end is
A)$\frac{4 \mathrm{~W}_{1}+3 \mathrm{~W}}{4 \mathrm{~A}}$
B)$\frac{3 \mathrm{~W}_{1}-4 \mathrm{~W}}{2 \mathrm{~A}}$
C)$\frac{3 \mathrm{~W}_{1}+4 \mathrm{~W}}{2 \mathrm{~A}}$
D)$\frac{4 \mathrm{~W}_{1}-3 \mathrm{~W}}{4 \mathrm{~A}}$

Correct option is A. $\text { stress }=\frac{\text { Tension }}{\text { Area }}$ Tension at height $\frac{3 \mathrm{~L}}{4}$ from lower end $\text { is } \frac{3}{4} \mathrm{w}+\mathrm{w}_{1}$ So,...

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A small mass ‘ $\mathrm{m}$ ‘ is suspended at the end of a wire having (negligible mass) length ‘L’ and cross-sectional area ‘ $A$ ‘. The frequency of oscillation for the S.H.M. along the vertical line is [ $\mathrm{Y}=$ Young’s modulus of material of the wire ]
A)$\frac{1}{2 \pi}\left[\frac{\mathrm{YAL}}{\mathrm{m}}\right]^{\frac{1}{2}}$
B)$\frac{1}{2 \pi}\left[\frac{\mathrm{YA}}{\mathrm{mL}}\right]^{\frac{1}{2}}$
C)$\frac{1}{2 \pi}\left[\frac{\mathrm{mA}}{\mathrm{YL}}\right]^{\frac{1}{2}}$
D)$\frac{1}{2 \pi}\left[\frac{\mathrm{YL}}{\mathrm{mA}}\right]^{\frac{1}{2}}$

Correct option is B. Let $F$ be the force applied on the end of wire and $x$ be the extension in the length of wire, then longitudinal strain $=\mathrm{x} / \mathrm{L}$ Normal stress $=$ F/A Young's...

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A body of mass $\cdot \mathrm{m}$ ‘ is moving along a circle of radius ‘ $r$ ‘ with linear speed ‘v’. Now, to change the linear speed to $\frac{V}{2}$ and to move it along the circle of radius ‘ $4 r^{\prime}$, required change in the centripetal force of the body is
A) decrease by $\frac{15}{16}$
B) decrease by $\frac{5}{16}$
C) increase by $\frac{9}{16}$
D) increase by $\frac{11}{16}$

Correct answer is A. $\begin{array}{l} F_{1}=\frac{m v^{2}}{r} \\ F_{2}=\frac{m}{4 r} \cdot \frac{v^{2}}{4}=\frac{1}{16} \frac{m v^{2}}{r}=\frac{F_{1}}{16} \\ \therefore...

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The electric field intensity at a point near and outside the surface of a charged conductor of any shape is ${ }^{\prime} E_{1}$ ‘. The electric field intensity due to uniformly charged infinite thin plane sheet is ‘ $\mathrm{E}_{2}$ ‘. The relation between ${ }^{\prime} \mathrm{E}_{1}$ ‘ and ‘ $\mathrm{E}_{2}$ ‘ is:
A) $2 \mathrm{E}_{1}=\mathrm{E}_{2}$
B) $\quad E_{1}=E_{2}$
C) $\quad E_{1}=2 E_{2}$
D) $\quad E_{1}=4 E_{2}$

Correct option is C. Electric field outside the conductor $\mathrm{E}_{1}=\frac{\sigma}{\epsilon_{0}}$ Electric field due to uniformly charged infinitely thin plate $\mathrm{E}_{2}=\frac{\sigma}{2...

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The relation between force ‘ $F$ ‘ and density ‘ $\mathrm{d}$ ‘ is $\mathrm{F}=\frac{\mathrm{x}}{\sqrt{\mathrm{d}}}$. The dimensions of $\mathrm{x}$ are
A) $\left[\mathrm{L}^{-1 / 2} \mathrm{M}^{3 / 2} \mathrm{~T}^{-2}\right]$
B) $\left[\mathrm{L}^{-1 / 2} \mathrm{M}^{1 / 2} \mathrm{~T}^{-2}\right]$
C) $\left[\mathrm{L}^{-1} \mathrm{M}^{3 / 2} \mathrm{~T}^{-2}\right]$
D) $\left[\mathrm{L}^{-1} \mathrm{M}^{1 / 2} \mathrm{~T}^{-2}\right]$

Correct option is A. $\text { The dimensions of } \mathrm{x} \text { are...

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The electron in the hydrogen atom is moving with a speed of $2 \times 10^{6} \mathrm{~m} / \mathrm{s}$ in an orbit of radius $0.5 \AA$. The magnetic moment of the revolving electron is
A)$15 \times 10^{-24} \mathrm{Am}^{2}$
B)$11 \times 10^{-24} \mathrm{Am}^{2}$
C)$6 \times 10^{-24} \mathrm{Am}^{2}$
D)$8 \times 10^{-24} \mathrm{Am}^{2}$

Correct option is D. $\text { Magnetic moment, } \mathrm{m}_{0}=\frac{\text { evr }}{2}$ $=\frac{1.6 \times 10^{-19} \times 2\times 10^{6} \times 0.5 \times 10^{-10}}{2}=8 \times 10^{-24}...

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A mass ‘ $\mathrm{m}$ ‘ suspended from a spring stretches it by $5 \mathrm{~cm}$ when on the surface of the earth . The mass is then taken on to a height of $1600 \mathrm{~km}$ above earth’s surface and again suspended from the same spring. At this altitude the extension of the spring is $\quad$ (Radius of earth $=6400 \mathrm{~km}$ )
A)$3.2 \mathrm{~cm}$
B) $1.6 \mathrm{~cm}$
C) $6.4 \mathrm{~cm}$
D)$0.8 \mathrm{~cm}$

Correct option is A. $\begin{array}{l} g=\frac{G M}{r^{2}} \\ \therefore \frac{g^{\prime}}{g}=\frac{r^{2}}{r^{2}} \\ R=6400 \mathrm{~km}, r^{\prime}=6400+1600=8000 \mathrm{~km} \\...

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For a photocell, the work function is ‘ $\phi$ ‘ and the stopping potential is ‘ $\mathrm{V}_{\mathrm{s}}$ ‘. The wavelength of the incident radiation is
A)$\frac{\mathrm{hc}}{\phi+\mathrm{eVs}}$
B)$\frac{\phi+\mathrm{eVs}}{\mathrm{hc}}$
C)$\frac{\phi-\mathrm{eVs}}{\mathrm{hc}}$
D)$\frac{\mathrm{hc}}{\phi-\mathrm{eVs}}$

Correct option is A. $\begin{array}{l} h v=\phi+\frac{1}{2} m v_{\max }^{2} \text { or } \frac{h c}{\lambda}=\phi+e V_{s} \\ \therefore \lambda=\frac{h c}{\phi+e V_{s}} \end{array}$

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A solenoid has core of a material with relative permeability 500 and its windings carry a current of $1 \mathrm{~A}$. The number of turns of the solenoid is 500 per metre. The magnetization of the material is nearly.
A) $\quad 2.5 \times 10^{3} \mathrm{~A} \mathrm{~m}^{-1}$
B) $2.5 \times 10^{5} \mathrm{Am}^{-1}$
C)$\quad 2.0 \times 10^{3} \mathrm{Am}^{-1}$
D)$2.0 \times 10^{5} \mathrm{Am}^{-1}$

Correct option is B) $2.5 \times 10^{5} \mathrm{Am}^{-1}$ Here, $\mathrm{n}=500$ turns $/ \mathrm{m}$ $\mathrm{I}=1 \mathrm{~A}, \mu_{\mathrm{r}}=500$ Magnetic intensity, $\mathrm{H}=\mathrm{nI}=500...

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Water rises in a capillary tube to a certain height such that the upward force due to surface tension is balanced by $63 \times 10^{-4} \mathrm{~N}$ force due to the weight of the water. The surface tension of water is $7 \times 10^{-2} \mathrm{~N} / \mathrm{m}$. The inner diameter of the capillary tube is nearly $(\pi=22 / 7)$
A)$6.3 \times 10^{-1} \mathrm{~m}$
B)$3 \times 10^{-2} \mathrm{~m}$
C)$7 \times 10^{-2} \mathrm{~m}$
D)$9 \times 10^{-2} \mathrm{~m}$

Correct answer is B. Upwards force due to surface tension, $F_{u}=(2 \pi R) \times T$ where, $2 \pi R \rightarrow$ circumference of tube $F_{u}=(2 \pi R) \times T$ Because surface tension is force...

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The moment of inertia of a ring about an axis passing through the centre and perpendicular to its plane is ‘I’. It is rotating with angular velocity $^{\prime} \omega^{\prime} .$ Another identical ring is gently placed on it so that their centres coincide. If both the rings are rotating about the same axis then loss is kinetic energy is
A) $\frac{\mathrm{I} \omega^{2}}{2}$
B) $\frac{\mathrm{I} \omega^{2}}{4}$
C) $\frac{\mathrm{I} \omega^{2}}{6}$
D) $\frac{\mathrm{I} \omega^{2}}{8}$

Correct option is B $\frac{\mathrm{I} \omega^{2}}{4}$ Initial angular momentum, $\mathrm{L}_{1}=\mathrm{I} \omega$ $\mathrm{K}_{1}=\frac{1}{2} \mathrm{I} \omega^{2}$ When, second ring is put on it...

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A particle of mass $\mathrm{m}$ is rotating in a plane in circular path of radius $\mathrm{r}$. Its angular momentum is $\mathrm{L}$. The centripetal force acting on the particle is.
A) $\frac{\mathrm{L}^{2}}{\mathrm{mr}}$
B) $\frac{\mathrm{L}^{2} \mathrm{~m}}{\mathrm{r}}$
C) $\frac{\mathrm{L}^{2}}{\mathrm{~m}^{2} \mathrm{r}^{2}}$
D) $\frac{\mathrm{L}^{2}}{\mathrm{mr}^{3}}$

Correct option is D) $\frac{\mathrm{L}^{2}}{\mathrm{mr}^{3}}$ Centripetal force, $F=\frac{\mathrm{mv}^{2}}{\mathrm{r}}=\frac{\mathrm{m}}{\mathrm{r}} \frac{\mathrm{L}^{2}}{\mathrm{~m}^{2}...

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The mutual inductance between two planar concentric rings of radii $r_{1}$ and $r_{2}\left(\right.$ with $\left.r_{1}>>r_{2}\right)$ placed in air is given by
A) $\frac{\mu_{0} \pi r_{2}^{2}}{2 r_{1}}$
B) $\frac{\mu_{0} \pi\left(r_{1}+r_{2}\right)^{2}}{2 r_{1}}$
C) $\frac{\mu_{0} \pi\left(r_{1}+r_{2}\right)^{2}}{2 r_{2}}$
D) $\frac{\mu_{0} \pi r_{1}^{2}}{2 r_{2}}$

The correct option is $\mathbf{A}$$\frac{\mu_{0} \pi r_{2}^{2}}{2 r_{1}}$ Magnetic field due to the larger coil at its centre is $B=\frac{\mu_{0} I}{2 r_{1}}$ where $\mathrm{I}$ is the current in...

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A resistance wire connected in the left gap of a metre bridge balances a $10 \Omega$ resistance in the right gap at a point which divides the bridge wire in the ratio $3: 2$. If the length of the resistance wire is $1.5 m$, then the length of $1 \Omega$ of the resistance wire is:
(1) $1.5 \times 10^{-1} m$
(2) $1.5 \times 10^{-2} m$
(3) $1.0 \times 10^{-2} m$
(4) $1.0 \times 10^{-1} m$

Correct option: (4) Explanation: Using the given ratio of wire, we have, $x \times 2 \lambda^{3}=10 \times 3 \lambda$ $x=15 \Omega$ $\therefore 15 \Omega \rightarrow 1.5 m$ $\therefore 1 \Omega...

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The capacitance of a parallel plate capacitor with air as medium is $6 \mu F$. With the introduction of a dielectric medium, the capacitance becomes $30 \mu F$. The permittivity of the medium is : $\left(\varepsilon_{0}=8.85 \times 10^{-12} C ^{2} N ^{-1} m ^{-2}\right)$
(1) $0.44 \times 10^{-10} C ^{2} N ^{-1} m ^{-2}$
(2) $5.00 C ^{2} N ^{-1} m ^{-2}$
(3) $0.44 \times 10^{-13} C ^{2} N ^{-1} m ^{-2}$
(4) $1.77 \times 10^{-12} C ^{2} N ^{-1} m ^{-2}$

Correct option: (1) Explanation: Given: $C_{\text {air }}=6 \mu F , C_{\text {medium }}=30 \mu F$ As we know, $C_{\text {air }}=\frac{\varepsilon_{0} A}{d}, C_{\text {medium }}=\frac{\varepsilon_{0}...

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An iron rod of susceptibility 599 is subjected to a magnetising field of $1200 A m ^{-1} .$ The permeability of the material of the rod is : $\left(\mu_{0}=4 \pi \times 10^{-7} T mA ^{-1}\right.$ )
(1) $2.4 \pi \times 10^{-5} T m A ^{-1}$
(2) $2.4 \pi \times 10^{-7} T m A ^{-1}$
(3) $2.4 \pi \times 10^{-4} T m A ^{-1}$
(4) $8.0 \times 10^{-5} T m A ^{-1}$

Correct option (3) Explanation: Given: $x _{ m }=599, \mu_{0}=4 \pi \times 10^{-7}$ $H =1200 A / m , \mu=?$ As we know, $\mu=\mu_{0}\left(1+ x _{ m }\right)$ Now, $=4 \pi \times 10^{-7}(1+599)$ $=4...

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Two cylinders A and B of equal capacity are connected to each other via a stop cock. A contains an ideal gas at standard temperature and pressure. B is completely evacuated. The entire system is thermally insulated. The stop cock is suddenly opened. The process is :
(1) isochoric
(2) isobaric
(3) isothermal
(4) adiabatic

Correct option (4) When cock is removed, dQ = 0 for the thermally insulated system. Because dw = 0 in vacuum Walls that are thermally insulated du + dw = dQ So du also a zero.

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When a ray of light is incident normally on one refracting surface of an equilateral prism of refractive index 1.5, the emerging ray $\left[\sin ^{-1}\left(\frac{1}{1.5}\right)=41.8^{\circ}\right]$
A) just grazes the second refracting surface.
B) is deviated by $20^{\circ}$.
C) is deviated by $30^{\circ}$.
D) undergoes total internal reflection at second refracting surface.

Correct answer is D. Critical angle for the material of prism $C=\sin ^{-1}\left(\frac{1}{\mu}\right)$ $=\sin ^{-1}=42^{\circ}$ since angle of incidence at surface $A B\left(60^{\circ}\right)$ is...

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When a resistance ‘ $R_{1}{ }^{\prime}$ is connected across the terminal of a cell of e.m.f. ‘ $\mathrm{E}_{1}$ ‘ the current is ‘ $\mathrm{I}_{1}$ ‘. When the resistance is changed to $\mathrm{R}_{2}$ ‘ the current is ‘ $I_{2}$ ‘. The internal resistance of the cell is
A. $\frac{I_{1} R_{2}+I_{2} R_{1}}{I_{1}+I_{2}}$
B. $\frac{I_{2} R_{2}+I_{1} R_{1}}{I_{1}+I_{2}}$
C. $\frac{I_{2} R_{2}-I_{1} R_{1}}{I_{1}-I_{2}}$
D. $\frac{I_{1} R_{2}-I_{2} R_{1}}{I_{1}-I_{2}}$

Correct answer is C. $\begin{array}{l} I_{1}=\frac{E}{r+R_{1}} \\ I_{2}=\frac{E}{r+R_{2}} \\ \frac{I_{1}}{I_{2}}=\frac{r+R_{2}}{r+R_{1}} \\ I_{1} r+I_{1} R_{1}=I_{2} r+I_{2} R_{2} \\...

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Three vessels A, B, C of different shapes have same base area and are filled with water up to same height ‘ $\mathrm{h}$ ‘. The respective forces exerted by water on the bases are ‘ $F_{A}^{\prime}, ‘ F_{B}^{\prime}, ‘ F_{c}^{\prime}$ and respective weights are ${ }^{\prime} \mathrm{W}_{\mathrm{A}}^{\prime},^{\prime} \mathrm{W}_{\mathrm{e}}^{\prime}, \mathrm{‘} \mathrm{W}_{\mathrm{c}}^{\prime}$. Then
A. $\quad \mathrm{F}_{\mathrm{A}}<\mathrm{F}_{\mathrm{B}}<\mathrm{F}_{\mathrm{C}} ; \mathrm{W}_{\mathrm{A}}<\mathrm{W}_{\mathrm{B}}>\mathrm{W}_{\mathrm{C}}$
B. $\quad F_{A}=F_{B}=F_{C} ; W_{A}>W_{B}>W_{C}$
C. $\quad F_{A}=F_{B}=F_{C} ; W_{A}D. $\quad \mathrm{F}_{\mathrm{A}}>\mathrm{F}_{\mathrm{B}}>\mathrm{F}_{\mathrm{C}} ; \mathrm{W}_{\mathrm{A}}>\mathrm{W}_{\mathrm{B}}<\mathrm{W}_{\mathrm{C}}$

Correct answer is B.

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The sequence of harmonics of a pipe open at one end and closed at the other end is $250 \mathrm{~Hz}$ and $350 \mathrm{~Hz}$. The resonating length of the air column in its fundamental mode will be (velocity of sound in air $=340 \mathrm{~m} / \mathrm{s})$
A. $1.8 \mathrm{~m}$
B. $1.6 \mathrm{~m}$
C. $1.7 \mathrm{~m}$
D. $1.4 \mathrm{~m}$

Correct answer is C. $\begin{array}{l} 2 n=n_{2}-n_{1} \\ =350-250=100 \\ n=\frac{100}{2}=50 H z \\ n=\frac{v}{4 l_{0}}=\frac{340}{4 \times l_{0}} \\ \therefore l=\frac{340}{4 \times...

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Two long, straight wires are set parallel to each other. Both the wires carry a current ‘I’ in opposite directions and separation between them is ‘ $2 \mathrm{R}$ ‘. The magnetic induction at the midway between them is $\left(\mu_{0}=\right.$ permeability of free space)
A. Zero
B. $\frac{\mu_{0} I}{4 \pi r}$
C. $\frac{\mu_{0} I}{\pi r}$
D. $\frac{\mu_{0} I}{2 \pi r}$

Correct answer is A. $\begin{array}{l} \mathrm{x}=\frac{\mathrm{r}}{2} \\ \mathrm{~B}_{1}=-\frac{\mu_{0} \mathrm{i}}{2 \pi \mathrm{x}} \\ \mathrm{B}_{2}=\frac{\mu_{0} \mathrm{i}}{2...

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A charged capacitor has charge ‘ $\mathrm{Q}_{1}$ ‘ at potential ${ }^{\prime} \mathrm{V}_{1}$ ‘. The charge on capacitor is increased to ‘ $\mathrm{Q}_{2}$ ‘ so that its potential increases to $\mathrm{V}_{2}$, then the difference between final and initial energy stored in capacitor is
A. $\frac{1}{2}\left(Q_{2}-Q_{1}\right)\left(V_{2}-V_{1}\right)$
B. $\frac{1}{2}\left(Q_{2} V_{2}-Q_{1} V_{1}\right)$
C. $\frac{1}{2}\left(\frac{Q_{2}^{2}}{V_{2}}-\frac{Q_{1}^{2}}{V_{1}}\right)$
D. $\frac{1}{2}\left(Q_{2} V_{2}^{2}-Q_{1} V_{1}^{2}\right)$

Correct answer is B.

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In Biprism experiment the distance between the two virtual images of the slits in the magnified and diminished position are $2.4 \mathrm{~mm}$ and $0.6 \mathrm{~mm}$ respectively. The distance between two coherent sources is
A. $1 \mathrm{~mm}$
B. $3.0 \mathrm{~mm}$
C. $1.2 \mathrm{~mm}$
D. $2.4 \mathrm{~mm}$

Correct answer is C. When, 1) magnified, distance between images, $\mathrm{d}_{1}=2.4$ 2) dimnished, distance between images, $\mathrm{d}_{2}=0.6$ $\therefore$ Distance between virtual sources...

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A sonometer wire of length ‘ $\mathrm{L}_{1}$ ‘ is in unison with a tuning fork of frequency’n’. When the vibrating length of the wire is reduced to ‘ $\mathrm{L}_{2}$ ‘, it produces ‘ $\mathrm{x}$ ‘ beats per second with the fork. The frequency of the fork is
A. $\frac{L_{2} x}{L_{2}-L_{1}}$
B. $\frac{L_{1} x}{L_{1}-L_{2}}$
C. $\frac{L_{2} x}{L_{2}-L_{1}}$
D. $\frac{L_{2} x}{L_{1}-L_{2}}$

Correct answer is D. For the length $L_{1}, n_{1}=\frac{1}{2 L_{1}} \sqrt{\frac{T}{m}}=n \ldots(1)$ where $m=$ mass per unit length For the length $L_{2}, n_{2}=\frac{1}{2 L_{2}} \sqrt{\frac{T}{m}}...

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A particle performs simple harmonic motion from mean position, with period $8 \mathrm{~s}$. The distance travelled by it between 1 st and 2 nd second of its motion is (A = amplitude of S. H. M.)
A. $A \frac{1}{2}$
B. $A\left(\frac{1}{\sqrt{2}}-1\right)$
C. $A\left(1-\frac{1}{\sqrt{2}}\right)$
D. $A\left(\frac{1}{\sqrt{2}}\right)$

Correct answer is C. In $2 \mathrm{~s}$ (which is equal to $\frac{T}{4}$ ), one amplitude will be convered. In $1 \mathrm{st}$ second $x=a \sin \left(\frac{\pi}{4}\right)=\frac{a}{\sqrt{2}}$...

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The rate of cooling of a body is $2{ }^{0} \mathrm{C} / \mathrm{min}$ when the body is at $50{ }^{0} \mathrm{C}$ above the temperature of surroundings. When the body is at $30^{\circ} \mathrm{C}$ above the temperature of surroundings, its rate of cooling will be
A. $0.6^{0} \mathrm{C} / \mathrm{min}$
B. $0.9^{0} \mathrm{C} / \mathrm{min}$
C. $1.2^{0} \mathrm{C} / \mathrm{min}$
D. $1.5^{0} \mathrm{C} / \mathrm{min}$

Correct answer is C. Rate of cooling $\frac{\mathrm{dT}}{\mathrm{dt}}=-\frac{\mathrm{k}}{\mathrm{ms}}\left(\mathrm{T}-\mathrm{T}_{\mathrm{s}}\right)$ Or $2=\frac{-\mathrm{k}}{\mathrm{ms}}(50)$ Or...

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The mass and radius of earth is $\mathrm{M}_{\mathrm{c}}$, and $\mathrm{R}_{\mathrm{c}}$, respectively and that of moon is $\mathrm{M}_{\mathrm{m}}$ and $\mathrm{R}_{\mathrm{m}}$ respectively. The distance between the centre of earth and that of moon is ‘D’ . The minimum speed required for a body (mass ‘m’) to project from a point midway between their centres to escape to infinity is
A. $\left(\frac{G D}{M_{e}}\right)^{1 / 2}$
B. $\sqrt{\frac{\left(M_{e}+M_{m}\right)}{D}}$
C. $2 \sqrt{G\left(M_{e}+M_{m}\right) / D}$
D. $\frac{G\left(M_{e}+M_{m}\right)}{2 D}$

Correct answer is C. From conservation of energy, since the velocity will be o at infinity, $\frac{1}{2} \mathrm{mv}^{2}-\frac{\mathrm{GmM}_{e}}{\mathrm{~d} / 2}-\frac{\mathrm{GmM}_{d}}{\mathrm{~d}...

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A point source is placed in air. The spherical wavefront has radius ‘ $r_{\mathrm{a}}$ ‘ after time ‘t’. If the same point source is placed in the medium of refractive index ‘ $\mu$ ‘, the radius of spherical wavefront in the medium in same time $t$ is
A. $\frac{r_{a}}{\mu}$
B. $\frac{r_{a}}{\mu^{2}}$
C. $\mu . r_{a}$
D. $\mu^{2} \cdot r_{a}$

Correct answer is A. Here Velocity of wave front $=\frac{R_{9}}{t}$ $\begin{array}{l} u=-1 \\ \frac{e}{R a / t}=1 \\ e=R a / t \end{array}$ in the meduim $u$ $\text { velochy }=R \frac{a}{t}$ so,...

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A solid sphere is rolling down a frictionless surface with translational velocity $\mathrm{V}$. It climbs the inclined plane from ‘A’ to ‘ $\mathrm{B}^{\prime}$ and then moves away from ‘ $\mathrm{B}^{\prime}$ on the smooth horizontal surface. The value of $V$ should be
A. $\sqrt{g h}$
B. $\geq\left[\frac{10 g h}{7}\right]^{1 / 2}$
C. $\sqrt{2 g h}$
D. $10 g h$

Correct answer is B. Applying law of conservation of energy for rotating body, $\begin{array}{l} \frac{1}{2} \mathrm{mv}^{2}+\frac{1}{2} \mathrm{I} \omega^{2}=\mathrm{mgh} \\ \frac{1}{2}...

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Glass has refractive index $\frac{3}{2}$ and water has refractive index $\frac{4}{3} .$ If the speed of light in glass is $2 \times 10^{8} \mathrm{~m} / \mathrm{s}$, then the speed of light in water will be
A. $1.6 \times 10^{8} \mathrm{~m} / \mathrm{s}$
B. $2.25 \times 10^{8} \mathrm{~m} / \mathrm{s}$
C. $1.8 \times 10^{8} \mathrm{~m} / \mathrm{s}$
D. $2 \times 10^{8} \mathrm{~m} / \mathrm{s}$

Correct answer is B. $\begin{array}{l} \quad \mathrm{v}_{\text {water }}=\frac{\mathrm{c}}{\mathrm{n}_{\text {water }}} \\ \therefore \quad \mathrm{v}_{\mathrm{water}}=\frac{3 \times 10^{8}}{(4 /...

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A transverse wave is travelling on a string with velocity ‘ $\mathrm{V}^{\prime} .$ The extension in the string is ‘ $\mathrm{x}^{\prime}$. If the string is extended by $50 \%$, then the speed of wave along the string will be nearly (Hooke’s law is obeyed)
A. $1.22 \mathrm{~V}$
B. $\frac{V}{1.5}$
C. $\frac{V}{1.22}$ D. $1.5 \mathrm{~V}$

Correct option is A. $\begin{array}{l} v=\sqrt{\frac{T}{m}} \\ \therefore \frac{\mathrm{V}_{2}}{\mathrm{~V}_{1}}=\sqrt{\frac{\mathrm{T}_{2}}{\mathrm{~T}_{1}}} \\ =\sqrt{1.5} \approx 1.22...

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A galvanometer of resistance $100 \Omega$ is connected to a battery of $2 \mathrm{~V}$ with a resistance of $1900 \Omega$ in series. The deflection obtained is 30 divisions. To reduce this deflection by 10 divisions the additional resistance required to be connected in series is
A. $1500 \Omega$
B. $500 \Omega$
C. $1000 \Omega$
D. $2000 \Omega$

Correct option is C. Total resistance in the circuit is $R=1900+100=2000 \Omega$ $\therefore$ Current $I_{1}=\frac{2}{2000}=10^{-3} A$ The deflection is decreased from 30 to 20 divisions $\therefore...

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When a certain photosensitive surface is illuminated with monochromatic light of frequency v, the stopping potential for the photocurrent is $\frac{\mathrm{V}_{0}}{2} .$ When the surface is illuminated by monochromatic light of frequency $\frac{\mathrm{v}}{2}$, the stopping potential is $-\mathrm{V}_{0}$. the threshold frequency for photoelectric emission is:
(A) $\frac{3 \mathrm{v}}{2}$
(B) $2 \mathrm{v}$
(C) $\frac{4}{3} \mathrm{v}$
(D) $\frac{5 \mathrm{v}}{3}$

Correct option is (A) $\frac{3 \mathrm{v}}{2}$ $\begin{array}{l} \mathrm{hv}=\mathrm{W}+\frac{\mathrm{v}_{0}}{2} \mathrm{e} \\ \frac{\mathrm{h} v}{2}=\mathrm{W}+\mathrm{v}_{0} \mathrm{e}...

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In balanced metre bridge, the resistance of bridge wire is $0.1 \Omega / \mathrm{cm}$. Unknown resistance ‘ $\mathrm{X}$ ‘ is connected in left gap and $6 \Omega$ in right gap, null point divides the wire in the ratio $2: 3$. Find the current drawn from the battery of $5 \mathrm{~V}$ having negligible resistance.
A) $1 \mathrm{~A}$
B) $1.5 \mathrm{~A}$
C) $2 \mathrm{~A}$
D) $5 \mathrm{~A}$

Answer is (A) $\frac{\ell_{1}}{\ell_{2}}=\frac{2}{3}=\frac{\mathrm{v}_{1}}{\mathrm{v}_{2}}$ Now $\quad \mathrm{v}_{1}+\mathrm{v}_{2}=\mathrm{v}=5 \mathrm{~V} \quad \mathrm{QV}=\mathrm{IR}$ For...

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Two particles $\mathrm{X}$ and $\mathrm{Y}$ having equal charges after being accelerated through same potential difference enter a region of uniform magnetic field and describe a circular paths of radii ‘ $\mathrm{r}_{1}$ ‘ and ‘ $r_{2}$ ‘ respectively. The ratio of the mass of $X$ to that of $Y$ is
A) $\frac{\mathrm{r}_{1}}{\mathrm{r}_{2}}$
B) $\sqrt{\frac{\mathrm{r}_{1}}{\mathrm{r}_{2}}}$
C) $\left[\frac{\mathrm{r}_{2}}{\mathrm{r}_{1}}\right]^{2}$
D) $\left[\frac{\mathrm{r}_{1}}{\mathrm{r}_{2}}\right]^{2}$

Answer is (A) The force acting on the particle inside magnetic field is $F_{B}=q v B \sin \theta$ This provides the necessary centripetal force $F_{c}=\frac{m v^{2}}{r}$ $\begin{array}{l} \therefore...

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When light of wavelength ‘ $\lambda$ ‘ is incident on photosensitive surface, the stopping potential is ‘ $\mathrm{V}$ ‘. When light of wavelength ‘ $3 \lambda$ ‘ is incident on same surface, the stopping potential is $\frac{\mathrm{V}^{\prime}}{6}$. Threshold wavelength for the surface is
A) $2 \lambda$
B) $3 \lambda$
C) $4 \lambda$
D) $5 \lambda$

Answer is (D) Using $\frac{\mathrm{hc}}{\lambda}-\phi=\mathrm{eV}$ When, wavelength $\lambda$ is incident on metallic surface, the stopping potential required to stop most energetic electron is...

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Two coherent sources ‘P’ and ‘Q’ produce interference at point ‘A’ on the screen where there is a dark band which is formed between $4^{\text {th }}$ bright band and $5^{\text {th }}$ bright band. Wavelength of light used is $6000 \AA$. The path difference between PA and QA is
A) $1.4 \times 10^{-4} \mathrm{~cm}$
B) $2.7 \times 10^{-4} \mathrm{~cm}$
C) $4.5 \times 10^{-4} \mathrm{~cm}$
D) $6.2 \times 10^{-4} \mathrm{~cm}$

Answer is (B) Here $\Delta \mathrm{x}=\frac{\lambda \delta}{2 \pi}$ For $n=4^{\text {th }}$ dark bone $\delta=(2 n+1) \pi \Rightarrow \delta=(8+1) \pi=9 \pi$ $\begin{array}{l} \Delta x=\frac{\lambda...

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A galvanometer of resistance $30 \Omega$ is connected to a battery of emf $2 V$ with $1970 \Omega$ resistance in series. A full scale deflection of 20 divisions is obtained in the galvanometer. To reduce the deflection to 10 divisions, the resistance in series required is
A) $4030 \Omega$
B) $4000 \Omega$
C) $3970 \Omega$
D) $2000 \Omega$

Answer is (C) $\mathrm{i}=\frac{\mathrm{V}}{\mathrm{R}_{\mathrm{eff}}}=\frac{\mathrm{V}}{1790+30}=\frac{2}{2000}=1 \times 10^{-3} \mathrm{~A}=1 \mathrm{~mA}$ This current provides full scale...

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Two identical parallel plate air capacitors are connected in series to a battery of e.m.f. ‘V’. If one of the capacitor is completely filled with dielectric material of constant ‘ $K$ ‘, then potential difference of the other capacitor will become
A) $\frac{\mathrm{K}}{\mathrm{V}(\mathrm{K}+1)}$
B) $\frac{\mathrm{KV}}{\mathrm{K}+1}$
C) $\frac{\mathrm{K}-1}{\mathrm{KV}}$
D) $\frac{\mathrm{V}}{\mathrm{K}(\mathrm{K}+1)}$

Answer is (B) When capacitors are connected is series $C_{e q}=\frac{C_{1} C_{2}}{C_{1}+C_{2}}$ and the potential areas plates of capacitors is given by $\mathrm{V}=\frac{\mathrm{Q}}{\mathrm{C}}$ If...

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The maximum frequency of transmitted radio waves above which the radio waves are no longer reflected back by ionosphere is – $\mathrm{N}=$ maximum electron density of ionosphere, $\mathrm{g}=$ acceleration due to gravity)
A) $\mathrm{gN}$
B) $\mathrm{gN}^{2}$
C) $g \sqrt{N}$
D) $\mathrm{g}^{2} \mathrm{~N}^{2}$

Answer is (C) For ionosphere, the maximum frequency of radio waves that can be reflected back is given by $g \sqrt{N}$. Where $N$ is maximum electron density of ionosphere and $g$ is acceleration...

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Light of wavelength ‘ $\lambda$ ‘ which is less than threshold wavelength is incident on a photosensitive material. If incident wavelength is decreased so that emitted photoelectrons are moving with same velocity then stopping potential will
A) increase
B) decrease
C) be zero
D) become exactly half

Answer is (A) According to photoelectric equation, $\frac{\mathrm{hc}}{\lambda}-\phi=\mathrm{E} \quad$ where $\mathrm{E}=\frac{\mathrm{p}^{2}}{2 \mathrm{~m}}(\mathrm{~K} . \mathrm{E})$ If $E$ is...

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Three parallel plate air capacitors are connected in parallel. Each capacitor has plate area $\frac{‘ A^{\prime}}{3}$ and the separation between the plates is ‘ $d$ ‘, ‘ $2 d$ ‘ and ‘ 3 d’ respectively. The equivalent capacity of combination is $\left(\epsilon_{0}=\right.$ absolute permittivity of free space)
A) $\frac{7 \in_{0} A}{18 \mathrm{~d}}$
B) $\frac{11 \in_{0} A}{18 \mathrm{~d}}$
C) $\frac{13 \in_{0} A}{18 \mathrm{~d}}$
D) $\frac{17 \in_{0} A}{18 \mathrm{~d}}$

Answer is (B) $\mathrm{C}=\mathrm{QV} \quad$ and $\mathrm{C}=\frac{\varepsilon_{0} \mathrm{~A}}{\mathrm{~d}}$ As given $\mathrm{C}_{1}=\frac{\varepsilon_{0} \cdot \mathrm{A}}{3 \mathrm{~d}} ; \quad...

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Interference fringes are produced on a screen by using two light sources of intensities ‘I’ and ‘9I’. The phase difference between the beams is $\frac{\pi}{2}$ at point $\mathrm{P}$ and $\pi$ at point $\mathrm{Q}$ on the screen. The difference between the resultant intensities at point $\mathrm{P}$ and $\mathrm{Q}$ is
A) $2 \mathrm{I}$
B) $4 \mathrm{I}$
C) $6 \mathrm{I}$
D) $8 \mathrm{I}$

Answer is (C) At point $\mathrm{P}$, resultant intensity of interfering wave is $\left(\mathrm{I}_{\mathrm{p}}\right)_{\mathrm{res}}=\mathrm{I}_{1}+\mathrm{I}_{2}+2 \sqrt{\mathrm{I}_{1}...

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An electron of mass ‘ $\mathrm{m}$ ‘ has de-Broglie wavelength ‘ $\lambda$ ‘ when accelerated through potential difference ‘ $\mathrm{V}^{\prime} .$ When proton of mass ‘ $\mathrm{M}$ ‘, is accelerated through potential difference ‘ $9 \mathrm{~V}$ ‘, the de-Broglie wavelength associated with it will be (Assume that wavelength is determined at low voltage)
A) $\frac{\lambda}{3} \sqrt{\frac{M}{m}}$B) $\frac{\lambda}{3} \cdot \frac{\mathrm{M}}{\mathrm{m}}$
C) $\frac{\lambda}{3} \sqrt{\frac{m}{M}}$
D) $\frac{\lambda}{3} \cdot \frac{m}{M}$

Answer is (C) When electron or any charged particle is accelerated through potential difference $v$, then kinetic energy gained is given by $E=e V$ $\begin{array}{l} E=\frac{1}{2} m...

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Alternating current of peak value $\left(\frac{2}{\pi}\right)$ ampere flows through the primary coil of the transformer. The coefficient of mutual inductance between primary and secondary coil is 1 henry. The peak e.m.f. induced in secondary coil is (Frequency of a.c. $=50 \mathrm{~Hz}$ )
A) $100 \mathrm{~V}$
B) $200 \mathrm{~V}$
C) $300 \mathrm{~V}$
D) $400 \mathrm{~V}$

Answer is (B) Peak value of current $I_{0}=1_{\operatorname{nms}} \times\sqrt{2}=\left(\frac{2}{\pi}\right)$ Amp Co-efficient of mutual inductance is $M=1$ horn Induced emf in secondary is given by...

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An iron rod is placed parallel to magnetic field of intensity $2000 \mathrm{~A} / \mathrm{m}$. The magnetic flux through the rod is $6 \times 10^{-4} \mathrm{~Wb}$ and its cross-sectional area is $3 \mathrm{~cm}^{2}$. The magnetic permeability of the rod in $\mathrm{Wb} / \mathrm{A}-\mathrm{m}$ is
A) $10^{-1}$
B) $10^{-2}$
C) $10^{-3}$
D) $10^{-4}$

Answer is (C) $\mu=\mathrm{B} \cdot \mathrm{H}=\frac{\phi}{\mathrm{A}} \times \mathrm{H}=\frac{6 \times 10^{-4}}{3 \times 10^{-4}} \times \frac{1}{2 \times 10^{3}}=10^{-3}$

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In potentiometer experiment, null point is obtained at a particular point for a cell on potentiometer wire $\mathrm{x} \mathrm{cm}$ long. If the length of the potentiometer wire is increased without changing the cell, the balancing length will (Driving source is not changed)
A) increase
B) decrease
C) not change
D) becomes zero

Answer is (A) For potentiometer, when null point is obtained for a particular cell (EV) at L $\mathrm{cm}$, (say). Whose length is $\mathrm{x} \mathrm{cm} \quad \therefore \mathrm{E}=\mathrm{L}...

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A simple pendulum of length ‘ $l$ ‘ has maximum angular displacement ‘ $\theta$ ‘. The maximum kinetic energy of the bob of mass ‘ $\mathrm{m}$ ‘ is (g = acceleration due to gravity)
A) $\operatorname{mg} l(1+\cos \theta)$
B) $\operatorname{mg} l\left(1+\cos ^{2} \theta\right)$
C) $\operatorname{mg} l(1-\cos \theta)$
D) $\mathrm{mg} l(\cos \theta-1)$

Answer is (C) When bob is at rest, the pendulum has only potential energy which is given as $\mathrm{P.E}=\mathrm{mg} \ell$. When bob displaces by small angler displacement $\theta$, the pendulum...

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Let ‘ $\mathrm{M}$ ‘ be the mass and ‘ $\mathrm{L}$ ‘ be the length of a thin uniform rod. In first case, axis of rotation is passing through centre and perpendicular to the length of the rod. In second case axis of rotation is passing through one end and perpendicular to the length of the rod. The ratio of radius of gyration in first case to second case is
A) 1
B) $\frac{1}{2}$
C) $\frac{1}{4}$
D) $\frac{1}{8}$

Answer is (B) M.I of rod whose axis of rotation is passing through center and perpendicular to the plane of rod is $\mathrm{I}=\frac{\mathrm{ML}^{2}}{12} \quad$ and $\quad...

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A black rectangular surface of area ‘A’ emits energy ‘E’ per second at $27^{\circ} \mathrm{C}$. If length and breadth are reduced to $\frac{1}{3}$ rd of initial value and temperature is raised to $327^{\circ} \mathrm{C}$ then energy emitted per second becomes
A) $\frac{4 \mathrm{E}}{9}$
B) $\frac{7 E}{9}$
C) $\frac{10 \mathrm{E}}{9}$
D) $\frac{16 \mathrm{E}}{9}$

Answer is (D) $\mathrm{E}=\mathrm{e} \sigma \cdot \mathrm{A}\left(\mathrm{T}^{4}-\mathrm{T}_{0}^{4}\right) \text { and } \mathrm{A}=\ell \mathrm{b}$ When $\ell$ and $b$ changes to $\frac{\ell}{3}$...

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Two particles of masses ‘ $\mathrm{m}$ ‘ and ‘ $9 \mathrm{~m}$ ‘ are separated by a distance ‘ $\mathrm{r}$ ‘. At a point on the line joining them the gravitational field is zero. The gravitational potential at that point is (G = Universal constant of gravitation)
A) $-\frac{4 \mathrm{Gm}}{\mathrm{r}}$
B) $-\frac{8 \mathrm{Gm}}{\mathrm{r}}$
C) $-\frac{16 \mathrm{Gm}}{\mathrm{r}}$
D) $-\frac{32 \mathrm{Gm}}{\mathrm{r}}$

Answer is (C) For the system described above $\mathrm{U}=\mathrm{m} \cdot \mathrm{V} \quad \therefore \mathrm{V}=\frac{\mathrm{U}}{\mathrm{m}}=\frac{-\mathrm{GM}}{\mathrm{X}}$ Here $M=16...

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A progressive wave is represented by $\mathrm{y}=12 \sin (5 \mathrm{t}-4 \mathrm{x}) \mathrm{cm}$. On this wave, how far away are the two points having phase difference of $90^{\circ}$?
A) $\frac{\pi}{2} \mathrm{~cm}$
B) $\frac{\pi}{4} \mathrm{~cm}$
C) $\frac{\pi}{8} \mathrm{~cm}$
D) $\frac{\pi}{16} \mathrm{~cm}$

Answer is (C) $y=12 m(5 t-4 y)$ Comparing in $y=A \sin (w t-k x)$ We have $A=12, w=5$ and $k=4$ $(w t-k x)=$ phase difference $=\frac{\pi}{2}$ $\therefore \quad 5 t-4 x=\frac{\pi}{2}$ When $t=0,4...

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Let a steel bar of length ‘ $l$ ‘, breadth ‘b’ and depth ‘d’ be loaded at the centre by a load ‘ $W$ ‘. Then the sag of bending of beam is ( $\mathrm{Y}=$ Young’s modulus of material of steel)
A) $\frac{\mathrm{Wl}^{3}}{2 \mathrm{bd}^{3} \mathrm{Y}}$
B) $\frac{\mathrm{W} l^{3}}{4 \mathrm{bd}^{3} \mathrm{Y}}$
C) $\frac{\mathrm{Wl}^{2}}{2 \mathrm{bd}^{3} \mathrm{Y}}$
D) $\frac{\mathrm{W} l^{3}}{4 \mathrm{bd}^{2} \mathrm{Y}}$

Answer is (B) The sag of bending of beam is given by. $\delta=\frac{W L^{3}}{48 \mathrm{YI}} \quad$ Where $W$ is load, $L$ is length, $Y$ is young's modules and $I$ is $\frac{b d^{3}}{12}$ (area...

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A disc of radius ‘ $R$ ‘ and thickness $\frac{\mathrm{R}}{6}$ has moment of inertia ‘I’ about an axis passing through its centre and perpendicular to its plane. Disc is melted and recast into a solid sphere. The moment of inertia of a sphere about its diameter is
A) $\frac{\mathrm{I}}{5}$
B) $\frac{\mathrm{I}}{6}$
C) $\frac{\mathrm{I}}{32}$
D) $\frac{\mathrm{I}}{64}$

Answer is (A) Volume of disc is $\mathrm{A} \cdot \mathrm{d}=\pi \cdot \mathrm{R}^{2} \times \frac{\mathrm{R}}{6}=\frac{\mathrm{R}^{3} \times \pi}{6}$ Moment of inertia of disc is...

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In a capillary tube of radius ‘ $R$ ‘, a straight thin metal wire of radius ‘ $r$ ‘ $(R>r)$ is inserted symmetrically and one end of the combination is dipped vertically in water such that the lower end of the combination is at same level. The rise of water in the capillary tube is $[\mathrm{T}=$ surface tension of water, $\rho=$ density of water, $\mathrm{g}=$ gravitational acceleration $]$
A) $\frac{\mathrm{T}}{(\mathrm{R}+\mathrm{r}) \rho \mathrm{g}}$
B) $\frac{\mathrm{R} \rho \mathrm{g}}{2 \mathrm{~T}}$
C) $\frac{2 T}{(R-r) \rho g}$
D) $\frac{(\mathrm{R}-\mathrm{r}) \rho \mathrm{g}}{\mathrm{T}}$

Answer is (C) $\mathrm{h}=\frac{2 \mathrm{~T} \cos \theta}{\rho g(\mathrm{R}-\mathrm{r})}$ For $\cos \theta=0, \mathrm{~h}=\frac{2 \mathrm{~T}}{\rho g(\mathrm{R}-\mathrm{r})}$

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The bob of a simple pendulum performs S.H.M. with period ‘ $\mathrm{T}$ ‘ in air and with period ‘ $\mathrm{T}_{1}$, in water. Relation between ‘ $\mathrm{T}$ ‘ and ‘ $\mathrm{T}_{1}$ ‘ is (neglect friction due to water, density of the material of the bob is $=\frac{9}{8} \times 10^{3} \mathrm{~kg} / \mathrm{m}^{3}$, density of water $=1 \mathrm{~g} / \mathrm{cc}$ )
A) $\mathrm{T}_{1}=3 \mathrm{~T}$
B) $\mathrm{T}_{1}=2 \mathrm{~T}$
C) $\mathrm{T}_{1}=\mathrm{T}$
D) $\mathrm{T}_{1}=\frac{\mathrm{T}}{2}$

Answer is (A) $\mathrm{T}=2 \pi \sqrt{\frac{\ell}{\mathrm{g}_{\mathrm{eff}}}}$ In air $\mathrm{g}_{\text {eff }}=\mathrm{g}$ where as in water $\begin{array}{l} g_{\text {eff...

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Two strings $\mathrm{A}$ and $\mathrm{B}$ of same material are stretched by same tension. The radius of the string A is double the radius of string B. Transverse wave travels on string A with speed ‘ $\mathrm{V}_{\mathrm{A}}$ ‘ and on string $B$ with speed ‘ $V_{B}$ ‘. The ratio $\frac{V_{A}}{V_{B}}$ is
A) $\frac{1}{4}$
B) $\frac{1}{2}$
C) 2
D) 4

Answer is (C) The velocity of wave travelling on string is $\begin{array}{l} v=n \lambda=\frac{\lambda}{2 L} \sqrt{\frac{T}{\mu}}=\sqrt{\frac{T}{\mu}} \\ \because \quad n=\frac{1}{2 L}...

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A particle moves along a circle of radius ‘r’ with constant tangential acceleration. If the velocity of the particle is ‘ $v$ ‘ at the end of second revolution, after the revolution has started then the tangential acceleration is
A) $\frac{v^{2}}{8 \pi r}$
B) $\frac{v^{2}}{6 \pi r}$
C) $\frac{v^{2}}{4 \pi r}$
D) $\frac{v^{2}}{2 \pi r}$

Answer is (A) Using $v^{2}-u^{2}=2 a s$ and $u=u_{1}$ and $s=4 \pi r$ $\therefore \quad 2 \mathrm{as}=\mathrm{v}^{2} \Rightarrow \frac{\mathrm{v}^{2}}{2...

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A mass ‘ $\mathrm{m}_{1}$ ‘ connected to a horizontal spring performs S.H.M. with amplitude ‘$A$’. While mass ‘ $\mathrm{m}_{1}$ ‘ is passing through mean position another mass ‘ $\mathrm{m}_{2}$ ‘ is placed on it so that both the masses move together with amplitude $\mathrm{A}_{1}$, The ratio of $\frac{\mathrm{A}_{1}}{\mathrm{~A}}$ is $\left(\mathrm{m}_{2}<\mathrm{m}_{1}\right)$
A) $\left[\frac{m_{1}}{m_{1}+m_{2}}\right]^{\frac{1}{2}}$
B) $\left[\frac{\mathrm{m}_{1}+\mathrm{m}_{2}}{\mathrm{~m}_{1}}\right]^{\frac{1}{2}}$
C) $\left[\frac{m_{2}}{m_{1}+m_{2}}\right]^{\frac{1}{2}}$
D) $\left[\frac{\mathrm{m}_{1}+\mathrm{m}_{2}}{\mathrm{~m}_{2}}\right]^{\frac{1}{2}}$

Answer is (A) P.E of the oscillating mass is given by $E=\frac{1}{2} m \omega x^{2}=\frac{1}{2} k x^{2} \quad \text { Where } k=m \omega^{2}$ When only black is oscillating, at $x=A$ $E=E_{\max...

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Assuming the expression for the pressure exerted by the gas on the walls of the container, it can be shown that pressure is
A) $[\frac{1}{3}]^{rd}$ kinetic energy per unit volume of a gas
B) $[\frac{2}{3}]^{rd}$ kinetic energy per unit volume of a gas
C) $[\frac{3}{4}]^{th}$ kinetic energy per unit volume of a gas
D) $\frac{3}{2}\times$ kinetic energy per unit volume of a gas

Answeris (B) The pressure exerted by the gas on the walls of container is $\mathrm{P}=\mathrm{P}_{0}+\mathrm{P}_{1}+\mathrm{P}_{2}$ i.e. $\quad \mathrm{P}=\mathrm{P}_{0}+\frac{1}{3} 8...

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When the observer moves towards the stationary source with velocity, ‘$V_1$’, the apparent frequency of emitted note is ‘$F_1$’. When the observer moves away from the source with velocity ‘$V_1$’, the apparent frequency is ‘$F_2$’. If ‘$V$’ is the velocity of sound in air and $\frac{F_1}{F_2}=2$ then $\frac{V}{V_1}=?$
A) 2
B) 3
C) 4
D) 5

Answer is (B) $\frac{\mathrm{f}_{1}}{\mathrm{f}_{2}}=2, \quad$ Speed of approach $=$ Speed of leaving The apparent frequency of sound level by observer when it is approaching source is given by...

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A series LCR circuit is connected to an AC voltage source. When $L$ is removed from the circuit, the phase difference between current and voltage is $\frac{\pi}{3}$. If instead $C$ is removed from the circuit, the phase difference is again $\frac{\pi}{3}$ between current and voltage. The power factor of the circuit is : (1) $1.0$ (2) $-1.0$ (3) zero (4) $0.5$

Correct option (1) Accoridng to question when inductor alone is removed, $\Delta \phi=\frac{\pi}{3}$ As we know that, $\tan \Delta \phi=\frac{X_{C}}{R}$ Thus, $\tan...

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In a quitar, two strings A and B made of same material are slightly out of tune and produce beats of frequency $6 \mathrm{~Hz}$. When tension in $\mathrm{B}$ is slightly decreased, the beat frequency increases to $7 \mathrm{~Hz}$. If the frequency of $\mathrm{A}$ is $530 \mathrm{~Hz}$, the original frequency of B will be :
(1) $536 \mathrm{~Hz}$
(2) 537 Hz
(3) $523 \mathrm{~Hz}$
(4) $524 \mathrm{~Hz}$

Correct option (4) Given, Frequency of $A=f_{a}=530 H z, f_{B}=?$ $\left|f_{A}-f_{B}\right|=6$, $\Rightarrow f_{B}<_{524}^{536}$ When Tension $\mathrm{B}$ decreases then $\mathrm{f}_{\mathrm{B}}$...

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A long solenoid of $50 \mathrm{~cm}$ length having 100 turns carries a current of $2.5 \mathrm{~A}$. The magnetic field at the centre of the solenoid is: $\left(\mu_{0}=4 \pi \times 10^{-7} \operatorname{Tm} \mathrm{A}^{-1}\right.$ )
(1) $6.28 \times 10^{-5} \mathrm{~T}$
(2) $3.14 \times 10^{-5} \mathrm{~T}$
(3) $6.28 \times 10^{-4} \mathrm{~T}$
(4) $3.14 \times 10^{-4} \mathrm{~T}$

Correct option is 3) Magnetic field for solenoid is given by formula $\mathrm{B}=\mu_{0} \mathrm{nI}$ $\mathrm{n}=\frac{\mathrm{N}}{\ell}$ Evaluating, $=4 \pi \times 10^{-7} \times 200 \times 2.5$...

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A capillary tube of radius $r$ is immersed in water and water rises in it to a height $h$. The mass of the water in the capillary is $5 \mathrm{~g}$. Another capillary tube of radius $2 r$ is immersed in water. The mass of water that will rise in this tube is :
(1) $10.0 \mathrm{~g}$
(2) $20.0 \mathrm{~g}$
(3) $2.5 \mathrm{~g}$
(4) $5.0 \mathrm{~g}$

Correct option: 1) As we know that, $h=\frac{2 T \cos \theta}{ rg }$ Given, $m =\rho \times$ volume $=\rho \times h \times \pi r ^{2}=5 gm$ According to information in the question, $r ^{\prime}=2...

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A short electric dipole has a dipole moment of $16 \times 10^{-9} \mathrm{C} \mathrm{m}$. The electric potential due to the dipole at a point at a distance of $0.6 \mathrm{~m}$ from the centre of the dipole, situated on a line making an angle of $60^{\circ}$ with the dipole axis is :
$\left(\frac{1}{4 \pi \varepsilon_{0}}=9 \times 10^{9} \mathrm{Nm}^{2} / \mathrm{C}^{2}\right)$
(1) $400 \mathrm{~V}$
(2) Zero
(3) $50 \mathrm{~V}$
(4) $200 \mathrm{~V}$

Correct option 4) Using the formula and evaluating, $\quad \mathrm{V}=\frac{\mathrm{kP} \cos \theta}{\mathrm{r}^{2}}=\frac{9 \times 10^{9} \times 16 \times 10^{-9} \times 1}{(0.6)^{2} \times...

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Match the corresponding entries of column 1 with column 2. [Where $\mathbf{m}$ is the magnification produced by the mirror]
A $\mathrm{A} \rightarrow \mathrm{b}$ and $\mathrm{c} ; \mathrm{B} \rightarrow \mathrm{b}$ and $\mathrm{c} ; \mathrm{C} \rightarrow \mathrm{b}$ and $\mathrm{d} ; \mathrm{D} \rightarrow \mathrm{a}$ and $\mathrm{d}$
B $\mathrm{A} \rightarrow \mathrm{a}$ and $\mathrm{c} ; \mathrm{B} \rightarrow \mathrm{a}$ and $\mathrm{d} ; \mathrm{C} \rightarrow \mathrm{a}$ and b; $\mathrm{D} \rightarrow \mathrm{c}$ and $\mathrm{d}$
C $\mathrm{A} \rightarrow \mathrm{a}$ and $\mathrm{d} ; \mathrm{B} \rightarrow \mathrm{b}$ and $\mathrm{c} ; \mathrm{C} \rightarrow \mathrm{b}$ and $\mathrm{d} ; \mathrm{D} \rightarrow \mathrm{b}$ and $\mathrm{c}$
D $\mathrm{A} \rightarrow \mathrm{c}$ and $\mathrm{d} ; \mathrm{B} \rightarrow \mathrm{b}$ and $\mathrm{d} ; \mathrm{C} \rightarrow \mathrm{b}$ and $\mathrm{c} ; \mathrm{D} \rightarrow \mathrm{a}$ and $\mathrm{d}$

Correct Option A Answer: The term "positive magnification" refers to an image that is upright in relation to the object. Negative magnification refers to an image that is inverted in relation to the...

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A car is negotiating a curved road of radius $\mathbf{r}$. The road is banked at an angle $\theta$. The coefficient of friction between the tyres of the car and the road is $\mu$ s. The maximum safe velocity on this road is:
A $\quad \sqrt{\operatorname{gR}^{2} \frac{\mu_{\mathrm{s}}+\tan \theta}{1-\mu_{\mathrm{s}} \tan \theta}}$
B $\sqrt{\operatorname{gR} \frac{\mu_{\mathrm{s}}+\tan \theta}{1-\mu_{\mathrm{s}} \tan \theta}}$
C $\sqrt{\frac{\mathrm{g}}{\mathrm{R}} \frac{\mu_{\mathrm{s}}+\tan \theta}{1-\mu_{\mathrm{s}} \tan \theta}}$
D $\sqrt{\frac{\mathrm{g}}{\mathrm{R}^{2}} \frac{\mu_{\mathrm{s}}+\tan \theta}{1-\mu_{\mathrm{s}} \tan \theta}}$

Correct Option B Solution: For vertical equilibrium on the curved road we have, $N \cos \theta=m g+f_{1} \operatorname{Sin} \theta----(1)$ Forhorizontal equilibrium on the curved road we have,...

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A small signal voltage $V(t)=V_{0} \sin \omega t$ is applied across an ideal capacitor C:
A Current I $(\mathrm{t})$, lags voltage $\mathrm{V}(\mathrm{t})$ by $90^{\circ}$
B Over a full cycle the capacitor $\mathrm{C}$ does not consume any energy from the voltage source.
C Current I (t) is in phase with voltage V (t).
D $\quad$ Current I $(\mathrm{t})$ leads voltage $\mathrm{V}(\mathrm{t})$ by $180^{\circ}$.

Correct Option B Solution: In AC circit the power is given by formula, $P=V I \operatorname{Cos} \phi$ Because the phase difference in a purely capacitive circuit is 90 degrees, no power is...

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