. What is the speed of light in glass? Speed of light in vacuum is 
.Ans: Given refractive index of glass, $\mu=1.5$ Speed of light, $\mathrm{c}=3 \times 10^{8} \mathrm{~m} / \mathrm{s}$ Now, we have the speed of light in glass given by the relation, $$...
Ans: We know that, $I=a^{2}+b^{2}+2 a b \cos \phi$ Where, $a$ and $b$ are amplitudes of two coherent waves that are having phase difference of $\phi$. Here $a^{2}=I, b^{2}=41$ $$ \begin{array}{l}...
Ans: Some differences between interference and diffraction are the following: \begin{tabular}{|l|l|} \hline \multicolumn{1}{|c|}{ Interference } & \multicolumn{1}{c|}{ Diffraction } \\ \hline...
Correct option is c $\frac{1}{2 \pi} \sqrt{\frac{1}{L C}-\frac{R^{2}}{L^{2}}}$ Megritude of admittanos $Y=\frac{1}{Z}$ is given by: $$ |\mathrm{Y}|=\frac{\mid \mathrm{H}^{2}...
The given statement is True.
Let an electric dipole consist of two equal and opposite point charges $-\mathrm{q}$ at $A$ and $+\mathrm{q}$ at $\mathrm{B}$,separated by a small distance $\mathrm{AB}=2 \mathrm{a}$, with centre at...
Consider an electric dipole $A B$ in which point charges $+q$ and $-q$ are separated by a distance 2l. Let $\mathrm{P}$ be a point on it's axis seperated by a distance r from the dipole's center...
emits maximum energyWien's constant $=2 \cdot 9 \times 10^{-3} \mathrm{~m}-\mathrm{K}$ Temperature at which maximum energy is radiated $\mathrm{T}=427^{\circ} \mathrm{C}$ $\therefore \mathrm{T}=427+273=700 \mathrm{~K}$...
A person with myopia cannot see distant objects clearly since the far point for that person is nearer than infinity. Such a person may see clearly up to a distance of a few meters. In this...
SOLUTION: Correct option is 4. Above the curie temperature the susceptibility of ferromagnetic substance varies inversely as the absolute temperature. Explanation: $\chi \propto...
is1 1.5 2 2.5 Solution: the correct answer is A. 1 $ \frac{{{\beta }_{dc}}-{{\alpha }_{dc}}}{{{\alpha }_{dc}}.{{\beta }_{dc}}}=\frac{\frac{\alpha }{1-\alpha }-\alpha }{\frac{\alpha }{1-\alpha }\times...
A. cos2ϕ/2 B. cos2ϕ/4 C. cos2ϕ/3 D. cos2ϕ Solution: The correct answer is A. cos2ϕ/2 The intensity of two similar light waves propagating in the same direction with a phase difference is as follows:...
Solution: The correct answer is B. $ We\,know\,that: $ $ \tan \theta =\frac{{{n}_{2}}}{{{n}_{1}}}\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,(1) $ $ {{n}_{1}}=1,\,\sin...
Solution: the correct answer is D. $ \frac{1}{{{C}_{initial}}}=\frac{1}{C}+\frac{1}{C} $ $ {{C}_{initial}}=\frac{C}{2} $ $ \frac{1}{{{C}_{final}}}=\frac{1}{KC}+\frac{1}{C} $ $...
400 W 800 W 600 W 200 W Solution: The correct answer is A 400 W Given : Vrms = 220 V, R = 18Ω and Z = 33Ω We get P = 400W
and its area of cross-section is 
. If 0.2 ampere current is flowing through the wire, the potential gradient of the wire is:10−1V/m 10−2V/m 10−3V/m 10−4V/m Solution: The correct answer is 2. 10−2V/m $ We\,have:\, $ $ \rho =40\times {{10}^{-8}}\,\Omega m\, $ $ I=0.2A $ $ A=8\times {{10}^{-6}}{{m}^{2}} $ $...
Both V and E increase Both V and E decrease V increases, E decreases V decreases, E increases Solution: The correct answer is B Both V and E decrease A dielectric capacitor with parallel plates....
λp ∝ λe λp ∝ λ2e λp ∝ √λe λp ∝ 1/λe Solution: The correct answer is A λp∝λe λe denotes the wavelength of the electron and λp denotes the wavelength of the proton. $ {{E}_{p}}=hv=\frac{hc}{{{\lambda...
(2) (1) (3) (4) Solution: The correct answer is At resonance ω =1/√LC f = 1/2π√LC The impedance is highest at resonance, and the frequency and current are lowest.
1.8 eV 1.4 eV 1 eV 1.267 eV Solution: The solution is D 1.267 eV According to the Einstein's Photoelectric equation:
. Its magnetic moment will be (cross – sectional area of a solenoid is very small as compared to its length, 
SI unit)A. 0.25 A B. 0.50 A C. 0.75 A D. 1 A Solution: the correct answer is C. 0.75 A Magnetic induction inside the solenoid is given by: $ B=\frac{{{\mu }_{0}}NI}{L} $ $ Magnetic\,Flux\,: $ $ \phi...
A. 2×105A/m B. 4×105A/m C. 3×105A/m D. 5×105A/m Solution: The correct answer is D. 5×105A/m $ Length\text{...
Repeater Modulation Attenuation Demodulation Solution: The correct answer is B. Modulation Modulation is the process of superimposing a signal on a high-frequency wave that functions as a carrier...
A. d2/D B. d2/2D C. d2/3D D. d2/4D Solution: The correct answer is C. d2/3D Because the second minima occur right in front of one of the slits, y = d/2 $ For\text{ }second\text{...
1.4 m Wb 2.2 m Wb 2.7 m Wb 2.9 m Wb Solution: The correct solution is C. 2.7 m Wb We have, di/dt = 10 M = E/(di/dt) =15/10 Flux = Mi = 1.5 × 1.8 Flux =2.7mWb
(A) which is always operated in reverse bias (B) which is always operated in forward bias (C) in which photo current is independent of intensity of incident radiation (D) which may be operated in...
A.1/n2 B. n2 C.1/n D. n Solution: The correct answer is D. n The magnetic moment of an electron can be expressed as: μ = −(neh)/(4πme) The magnetic moment is directly proportional to the primary...
2 eV 4 eV 8 eV 6 eV Solution: The correct answer is D 6 eV E = (−13.6/4) + (13.6/1) = 10.2 So stopping potential equals to hv = ϕ(0) + Vs 10.2 − 4.2 = Vs Vs = 6 eV
1.258 1.269 1.310 1.286 Solution: The correct answer is D 1.286 The number of waves on the glass slab equals the number of waves in the water column. μghg = μwhw μw = μghg/hw = 1.5 × 67 μw = 1.286...
A. 0.2 $ \log _{e}^{2} $ B. 0.5 $ \log _{e}^{2} $ C. 0.6 $ \log _{e}^{2} $ D. 0.8 $ \log _{e}^{2} $ Solution: The correct answer is 0.5 $ \log _{e}^{2} $ $ {{N}_{0}}\lambda =10000 $ $ N\lambda =2500...
A. 10−2 m B. 10−1 m C. 2×10−2 m D. 2×10−1 m Solution: The correct answer is A. 10−2 m expression for the distance between the first two minimum on the other side is given by: = 2λD/d...
12 24 48 36 Solution: The correct answer is B. 24 let the resistance at right be R and at left be r $ \frac{r}{R}=\frac{{{l}_{x}}}{{{l}_{R}}}=\frac{40}{60} $ $ \frac{r+30}{R}=\frac{60}{40} $ $...
s/3 s/6 s/12 s/9 Solution: The correct answer is D. s/9 S = G/8 S = G/(n−1) Now, we get: G/8 = G/(n−1) ⇒ n = 9 When a galvanometer's range is increased nine times, its sensitivity drops to 1/9. S'=...
5μF 7μF 6μF 8μF Solution; The correct answer is A. 5μF Let the capacitance be C when connected in parallel combination, C1 = C + C + C = 3C when other was connected in series, we get: 1/Cf = 1/3.75...
v1−v2=v3 v1+v3=v2 v1+v2=v3 v1−v3=2v1 Solution: The correct answer is A. v1−v2=v3 $ The\,Balmer\,\lim it\,wavelength\,is: $ $ \frac{1}{{{\lambda }_{1}}}=R\left( \frac{1}{{{2}^{2}}}-\frac{1}{\infty }...
A. MgA2/2L B. MgA/2L C. MgA2/L D. 2MgA2/L Solution: The correct answer is A. MgA2/2L $ PE=\frac{m{{\omega }^{2}}{{x}^{2}}}{2} $ $ \omega =\sqrt{\frac{g}{L}}\,at\,extreme\,po\operatorname{int}\,x=A $...
is conserved increases decreases may increase or decrease Solution: The correct answer is A. is conserved The kinetic energy of a particle travelling in a vertical circle is converted into potential...
100,200,300,400,..... 100,300,500,700,.... 200,300,400,500,..... 200,400,600,800,..... Solution: The correct answer is D. 200,400,600,800,..... $ Fundamental\text{ }frequency\text{ }of\text{...
0 rad π/4 rad π/2 rad π rad Solution: The correct answer is D. pi rad The following is the displacement of the particle executing SHM. x = A coswt Differentiating w.r.t. time we get v = dx/dt v =−...
Solution: The correct answer is B. The following is the expression for acceleration due to gravity at a depth of d below the earth's surface. $ g'={{g}_{s}}\left( 1-\frac{d}{R} \right) $ $...
0.4 0.2857 0.2257 0.3557 Solution: The correct answer is B. 0.2857 $ Universal\text{ }gas\text{ }constant\text{ }is\text{ }given\text{ by:} $ $ R=nCp $ $ For\text{ }diatomic\text{ }molecule,\text{...
1.5 cm 2 cm 2.5 cm 3 cm Solution: The correct answer is A. 1.5 cm $ Upward\text{ }force\,is:\,\,\,F=105~dyne~=105\times {{10}^{-5}}~N $ $ Surface\text{ }tension\text{ }of\text{ }water~T=7\times...
1/3 1/5 1/4 1/6 Solution: The correct answer is A. 1/3 $ Fundamental\text{ }frequency\text{ }of\text{ }the\text{ }first\text{ }wire~\,is: $ $...
96 rad 48 rad 24 rad 72 rad Solution: The correct answer is 96 rad Angular acceleration is given by: α = 24/8 = 3 rad/s2 Now angular displacement becomes: θ = 0 + αt2/2 θ...
√14 m √2.8 m √1.4 m √0.7 m Solution: The correct answer is B. √2.8 m $ We\,have:\,\,m=2kg~~and\,~v=6m/s $ $ Total\text{ }kinetic\text{ }energy\text{ }of\text{ }the\text{ }sphere\text{...
Solution: The correct answer is B. $ change\text{ }in\text{ }length:\,~\delta L=L\alpha T $ $ final\text{ }length~\,{{L}_{f}}=L+\delta L $ Now, in order to limit the expansion, there must be a force...
A. 2nv0/v B. v/2nv0 C. v/nv0 D. nv0/v Solution: The correct answer is A. 2nv0/v $ \text{Making use of the }doppler\text{ }effect $ $ Case\text{ }I:~Observer\,is\text{ }approaching\text{ }the\text{...
The correct option is B. $ u=\omega \sqrt{({{A}^{2}}-{{x}^{2}})} $ $ \alpha =-{{\omega }^{2}}x $ $ v=\omega \sqrt{({{A}^{2}}-{{y}^{2}})} $ $ \beta =-{{\omega }^{2}}y $ $...
Solution: The correct answer is D. $ Initial\text{ }angular\text{ }momentum\text{ }of\text{ }the\text{ }system~is\,given\,by: $ $ {{L}_{i}}={{I}_{1}}{{\omega }_{1}}+{{I}_{2}}\omega \prime ~ $ $...
8n/15 4n/15 16n/15 32n/15 Solution: The correct answer is C. 16n/15 $ {{(\omega /4)}^{2}}={{\omega }^{2}}-2\alpha ~n(2\pi ) $ $ \therefore 2\alpha n(2\pi )={{\omega }^{2}}-\frac{{{\omega }^{2}}}{16}...
Solution: The correct answer is B. $ Let\text{ }the\text{ }tension\text{ }be\text{ }F.\,So, $ $ F-mg=ma\to F=m(g+a) $ $ Now,\,~T=\frac{F}{\pi {{r}^{2}}} $ $ r={{\left( \frac{m\left( g+a \right)}{\pi...
8 14 20 17 Solution: The correct answer is D. 17 $ The\,fundamental\text{ }frequencyfor\,closed\,pipe\,is: $ $ {{f}_{c}}=\frac{v}{4l} $ $ for\,open\,pipe:\,\,{{f}_{o}}=\frac{v}{2l} $ $ \therefore...
T1/T2 (T1/T2)2 (T2/T1)2 T2/T1 Solution: The correct answer is 2. (T1/T2)2
Velocity V=1.5m/s Frequency F=0.2 Hz Wavelength λ=10 m Amplitude A=3 cm Solution: The correct answer is C. Wavelength λ=10 m $ Y=3\sin \left[ \pi \left( \frac{t}{3}-\frac{x}{5} \right)+\frac{\pi...
24 Kg m2/s 48 Kg m2/s 72 Kg m2/s 96 Kg m2/s Solution: The correct answer is C. 72 Kg m2/s $ {{w}_{f}}=w+\alpha t\to 0=60+\alpha \times 300 $ $ \alpha =\frac{-1}{5} $ $ now\text{ }to\text{...
L/6,L/2,5L/6 L/8,L/4,L/2 L/2,L/4,L/6 L/6,L/2,L/6 Solution: The correct answer is A. L/6,L/2,5L/6 $ For\text{ }second\text{ }overtone~:\frac{3\lambda }{2}=L $ $ first\text{ }maxima\text{ }will\text{...
2√2m 4√2m 6√2m 8√2m Solution: The correct answer is D. 8√2m $ We\,have~: $ $ x=A\,sin\omega t $ $ Upon\text{ }differentiating\text{ }both\text{ }side,\,we\,get: $ $ ~v=A\omega \,cos\omega t $ $...
2(R+h)/R (R+h)/2R (R+h)/R R/(R+h) Solution: The correct answer is 1. 2(R+h)/R
n2:1 n:1 3√n:1 √n:1 Solution: The correct answer is C. $ By\text{ }mass\text{ }conservation\text{ }new\text{ }radius\text{ }R\text{ }is~ $ $ R={{n}^{1/3}}r $ $ ratio\text{ }of\text{ }energy\text{...
Pm/KT KT/Pm Km/PT PK/Tm Solution: The correct answer is A. Pm/KT The ideal gas equation is PV = nRT, we can also write: PV = NKT here K is a Boltzmann constant Now we have: ρ = m/V = Nm/V putting...
0 4 5 6 Solution: The correct answer is A. 0 The frequency for a pipe closed at one end is given by: f = (n×v )/4L v = speed of sound = 332m/s L = length of pipe = 83cm = 0.83m n = 0,1,2... f =...
h=2R h=3R h=5R h=7R Solution: The correct answer is D. h=7R $ Escape\,velocity,\,{{v}_{e}}=\sqrt{\frac{2GM}{R}} $ $ v=\frac{{{v}_{e}}}{4}=\frac{1}{4}\sqrt{\frac{2GM}{R}} $ $ Centripetal\text{...
5 cm 20 cm 15 cm 10 cm Solution: The correct answer is D. 10cm We know that the magnetic field at a distance d from the magnet's centre on the axis is given as follows for a magnet of length 2l and...
F2L/2AY FL/2AY FL/AY F2L2/2AY Solution: The correct answer is 1. $ Energy,\,\,U=\frac{1}{2}\times stress\times strain\times volume $ $ Stress=\frac{F}{A} $ $ Strain=\frac{Stress}{Y}=\frac{F}{AY} $ $...
n √n n√2 n√5 Solution: The correct answer is A. n The number of waves is determined by the same harmonic. As a result, the needed number of waves is the same, n.
h2/mλ h2/2m2λ h2/2m2λ2 h2/2mλ2 Solution: $ \lambda =\frac{h}{mv} $ $ v=\frac{h}{m\lambda } $ $ K=\frac{1}{2}m{{v}^{2}}=\frac{1}{2}m{{\left( \frac{h}{m\lambda } \right)}^{2}} $ $...
(A) large and large (B) large and small (C) short and large (D) short and small Solution: The correct answer is (C) short and small Two convex lenses make up a compound microscope. The objective...
(1) (2) (3) (4) Solution: The correct answer is A. (1) The stopping potential is the amount of energy required to prevent an electron from reaching the outside world. Electrons are emitted from the...
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be the physical quantity. If the percentage error in the measurement of physical quantities a, b and c is 2,3 and 4 percent respectively then percentage error in the measurement of x is28% 7% 21% 14% Solution: The correct option is D. 14%
Solution: The correct answer is C. $ Y=\frac{F/A}{\Delta l/l} $ $ \Delta l=\frac{Fl}{AY} $ $ \Delta l\,\alpha \,\frac{1}{A} $ $ m=Al\rho \Rightarrow m\,\alpha \,A $ $ Hence,\,\frac{\Delta...
100J 500J 700J 900J Solution: The correct answer is 2. 500 J Let, the heat transmitted per minute be given by H. Now, The total incident energy = energy absorbed + energy reflected + energy...
Solution: The correct answer is A. $ For\,series\,combination,\,equivalence\,capacitance\,=\,\frac{{{C}_{1}}}{{{N}_{1}}} $ $ ch\arg e,\,\,q=(3V)\frac{{{C}_{1}}}{{{N}_{1}}} $ $...
Solution: The correct answer is D. $ We\,have:\,\,g=\frac{G{{M}_{e}}}{{{R}^{2}}} $ $ Applying\text{ }law\text{ }of\text{ }conservation\text{ }of\text{ }mechanical\text{ }energy:- $ $...
1/2 (1/3)rd (1/4)th (1/8.5)th Solution: The correct answer is The current through the galvanometer is given by $ {{i}_{1}}=\frac{{{R}_{s}}}{{{R}_{s}}+{{R}_{g}}}i $ where Rs = resistance of shunt...
(A) decreases while frequency increases (B) remains the same whereas frequency increases (C) increases and frequency also increases (D) decreases while frequency remains the same Solution; The...
independent of d directly proportional to d directly proportional to 1/√d directly proportional to √d Solution: The correct answer is A. independent of d The required electric field for the point...
g 3g 6g 4g Solution: The correct answer is B. 3g $ The\text{ }speed\text{ }when\text{ }a\text{ }string\text{ }is\text{ }horizontal,~\,v=\sqrt{3gr} $ $ Hence,\text{ }centripetal\text{...
(A) decreases (B) increases (C) becomes zero (D) first decreases and then increases Solution: The correct answer is (B) increases We know: eVo = hν − ϕ where Vo gives the stopping potential, ν is...
P √5P √11P √13P Solution: The correct answer is √13P P1 = −3Pi and P2 = 2Pj And m1 = m2 = m3 = m/3 And Pinitial = 0 Let the momentum of the third part be P3 Making use of the principle of...
Iω2/2 Iω2/4 Iω2/6 Iω2/8 Solution: The correct answer is Iω2/4 $ Initial\text{ }angular\text{ }momentum,\,{{L}_{1}}=I\omega $ $ {{K}_{1}}=\frac{1}{2}I{{\omega }^{2}} $ We know that the block's...
2πm/s πm/s 1/2πm/s 1/πm/s Solution: The correct answer is 1/πm/s We know that the block's greatest speed occurs when it is in balance. Applying mechanical energy conservation at the equilibrium...
(A) small, positive and small, positive (B) large, positive and small, negative (C) small, positive and small, negative (D) large, negative and large, positive Solution: The correct answer is (C)...
√0.6 0.4 0.2 √0.2 Solution: The correct answer is
2ml2/3 4ml2/3 5ml2/3 ml2/12 Solution: The correct answer is The inertia moments of rods AD and BC are as follows: $ \frac{1}{3}m{{T}^{2}} $ Because rod AB is on the X-axis, its moment of inertia...
Solution: The correct answer is A.
0.1m 0.2m 0.6m 0.8m Solution: The correct answer is There are six loops in the fifth overtone, each of which is half wavelength long. $ 6\frac{\lambda }{2}=L=2.4\,m $ $ \lambda =0.8\,m $ Now,...
4 cm 8 cm 12 cm 16 cm Solution: the correct answer is B. 8cm We know, $ v=\frac{c}{\mu }\,\,where\,\,\mu =1.6 $ Let T1 represent the time it takes to travel from the source to the surface, and T2...
Solution: The correct answer is C. $ Time,\,\,T=\frac{1}{y} $ $ speed\,of\,proton,\,v=\frac{2\pi R}{T}=2\pi Ry $ $ For\text{ }circular\text{ }motion\text{ }in\text{ }the\text{ }dees- $ $...
(A) is one in which width of depletion layer increases (B) is one in which potenital barrier increases (C) acts as closed switch (D) acts as open switch Solution: The correct answer is (C) acts as...
3h 4h 2h h Solution: The correct answer is A. 3h $ h=\frac{2T\cos \theta }{r\rho g} $ $ \Rightarrow h\,\alpha \,\frac{1}{r} $ $ A=\pi {{r}^{2}},\,here\,\,r\,\alpha \,\sqrt{A} $ $ \therefore...
Solution; The correct answer is D. $ \frac{{{C}_{p}}}{{{C}_{v}}}=\gamma $ $ {{C}_{p}}=\gamma {{C}_{v}} $ $ {{C}_{p}}-{{C}_{v}}=R $ $ \gamma {{C}_{v}}-{{C}_{v}}=R $ $ {{C}_{v}}=\frac{R}{\gamma -1}...
Solution: The correct answer is A. The resistance of a body is related to its length and material resistivity, and inversely proportional to its cross-sectional area. $ {{R}_{1}}=\rho...
1/3A 4/3A 5/3A 7/3A Solution; The correct answer is B. 4/3 A Suppose that there is no loss in power transmission: $ {{E}_{1}}{{I}_{1}}={{E}_{2}}{{I}_{2}} $ $...
2an 4an 8an 6an Solution; the correct answer is In SHM:- Distance covered per oscillation, s = 4a Time taken for one oscillation, T = 1/n Hence, the average speed per oscillation, v(avg) = S/T =...
(e−1)/2 (e+1)/2 e e/2 Solution: the correct answer is (e+1)/2 $ coefficient\text{ }of\text{ }restitution,\,\,e=\frac{{{v}_{2}}-{{v}_{1}}}{u} $ $...
Solution: The correct answer is B. $\sqrt{\frac{T}{\pi \rho g}} $ $ W={{F}_{s}} $ $ \pi {{r}^{2}}L\rho g=LT $ $ r=\sqrt{\frac{T}{\pi \rho g}} $
9Rh/16m 11hR/16m 13hR/16m 15hR/16m Solution: the correct answer is 4. 15hR/16m The change in energy is given by: $ \Delta E=Rhc\left( \frac{1}{n_{1}^{2}}-\frac{1}{n_{2}^{2}} \right) $ $ \Delta...
Solution: The correct answer is B. Mz = CB/T $ According\text{ }to\text{ }Curie's\text{ }law:- $ $ X\,\alpha \,\frac{1}{T'} $ $ \Rightarrow X=\frac{C}{T'} $ $ Magnetization,\,{{M}_{z}}=XB $ $...
(A) resolving power remains constant (B) resolving power becomes zero (C) limit of resolution is decreased (D) limit of resolution is increased Solution; The correct answer is (D) limit of...
1. α2r2 / v 2. α2r2 / v2 3. v2r2 / α 4. α2r/v2 Solution: The correct answer is 2. α2r2 / v2 $ Tangential\text{ }acceleration,~\text{ }~{{a}_{t}}=\alpha .r~ $ $ {{a}_{r}}=\frac{{{v}^{2}}}{r} $ $...
(A) glancing angle (B) angle of incidence (C) angle of deviation (D) angle of refraction Solution: The correct answer is (A) glancing angle The angle formed by the incident beam ray and the...
0.8A 1.0A 1.2A 1.4A Solution: The correct answer is In the circuit above, we have 15/3 = 20/4 As a result, the wheat-stone bridge is balanced, and the galvanometer has no current flowing through...
5 mA 20 mA 10 mA 15 mA Solution: The correct answer is 20mA $ E=200\sqrt{2}sin(100t) $ $ {{E}_{rms}}=\frac{{{E}_{0}}}{\sqrt{2}}=200V\,and\,\omega =100\,{{s}^{-1}} $ $ {{X}_{C}}=\frac{1}{{{\omega...
Solution: The correct answer is B. 2 × (I1+I2) since the frequencies are the same so the maximum intensity will be I = (√I1+√I2 )2 minimum intensity is i = (√I1-√I2 )2 adding both we get, I + I =...
R√3 2R 3R R√2 Solution: The correct answer is A. R√3 We know that the magnetic field B is given by at a distance x from the center of a current-carrying coil's axis. $ {{B}_{1}}=\frac{{{\mu...
A) MRω/4t B) MRω/2t C) MRω/t D) MRωt Solution: The correct answer is B) MRω/2t $ \alpha =\frac{\omega }{t} $ $ moment\text{ }of\text{ }inertia\text{ }of\text{ }disc,\,I=\frac{1}{2}M{{R}^{2}} $ $...
1000 1100 900 1200 Solution: The correct answer is A. 1000 $ \frac{{{V}_{0}}}{{{V}_{i}}}=\beta \frac{{{R}_{L}}}{{{R}_{I}}} $ $ \beta =\frac{{{I}_{C}}}{{{I}_{B}}}=\frac{1.5mA}{20\mu A}=75 $ $...
r.(π)1/4(3)1/4 r.(π)1/2(3)1/4 r12(π)1/3(3)1/2 r12(π)1/2(3)1/2 Solution: The correct answer is 2. r.(π)1/2(3)1/4 $ velocity\text{ }of\text{ }efflux\text{ }at\text{ }depth~d~is\,v=\sqrt{2gd} $ $...
A) 2πcm/s B) 4πcm/s C) 16πcm/s D) 8πcm/s Solution: the correct answer is D) 8πcm/s $ l=1m=100cm~and~s=16cm $ $ l\times (2\theta )=s $ $ 2\theta =\frac{s}{l}=0.16 $ $ \theta =0.08 $ Applying the Law...
A) decreasing the number of turns of coil B) increasing the number of turns of coil C) decreasing the area of a coil D) by using a weak magnet Solution: The answer is B) increasing the number of...
A) 2E1 = E2 B) E1 = E2 C) E1 = 2E2 D) E1 = 4E2 Solution: The answer is C) E1 = 2E2 The electric field...
A) i B) i/3 C) i/4 D) 3i/4 Solution; the answer is C) i/4 $ speed\text{ }of\text{ }light\text{ }in\text{ }medium\text{ }~v=\frac{3c}{4} $ $ Using~c\,sinr=v\,sini $ $ As~i~and~r~are\text{ }very\text{...
Solution; The answer is A) $ We\,know\,that:\,\,F=qE $ $ Thus\text{ }acceleration\text{ }of\text{ }the\text{ }electron~a=\frac{F}{m}=\frac{qE}{m} $ $ and\,here\,a=0 $ $ Using\text{...
A) AND B) OR C) NAND D) NOR Solution: The answer is D) NOR The truth table for the NOR gate is shown above, indicating that it is a logic gate that...
Solution: The answer is B) Assume the satellite is in a circular orbit with a radius of r and a linear velocity of v. The gravitational force of the earth balances the centripetal force acting on...
KTg/4π2 KT2g/4π2 KTg2/π2 KT2g2/π2 Solution: The answer is 2. KT2g/4π2 $ Time\text{ }period\text{ }of\text{ }the\text{ }suspended\text{ }mass\text{ }executing\text{ }SHM\,Is: $ $ T=2\pi...
Solution: The correct answer is (c) $ {{n}_{1}}=\frac{pv}{2\left( {{l}_{1}}+2e \right)}\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,(1) $ $ {{n}_{2}}=\frac{pv}{2\left( {{l}_{2}}+2e...
Solution: The correct answer is D) $ T=\frac{m{{v}^{2}}}{r} $ $ Time~Period=\frac{2\pi r}{v} $ $ ~\Delta L=\frac{Fl}{AY} $ $ ~\Rightarrow T=\frac{\Delta lAY}{l} $ $ v=\sqrt{\frac{\Delta lAYr}{lm}} $...
Solution: The correct answer is (c) $ Using\text{ }law\text{ }of\text{ }conservation\text{ }of\text{ }energy.\, $ $ At\text{ }height\text{ }h:~Mechanical\text{ }Energy=~M.{{E}_{1}}=mgh $ $...
A) more at T1 B) more at T2 C) equal for T1 and T2 D) independent of T1 and T2 Solution; the correct answer is B) more at T2 The expression for heat radiated per unit time is P = σAeT4 where σ...
Solution: The correct answer is (A) Let's say there are n little drops that have fused to produce one larger drop. Before and after the merging, the liquid's mass is conserved. $ \therefore \rho...
A) Rω/4S B) R2 ω2/4S C) Rω/2S D) R2 ω2/2S Solution: The correct answer is R2 ω2/4S $ Moment\text{ }of\text{ }inertia\text{ }of\text{ }hollow\text{ }sphere~I=\frac{2}{3}M{{R}^{2}} $ $ Kinetic\text{...
A) n/4 B) n/8 C) n/12 D) n/16 Solution: The correct answer is $f=\frac{1}{2L}\sqrt{\frac{T}{\mu }}$ According to the question, we have: T2 = T1, μ2 = 4μ1, L2 = 2L1 [∵ A2...
A) 2000 N B) 1000 N C) 500 N D) 250 N Solution: The correct answer is A) 2000N We know that tension in a rope is directly proportional to r2, thus doubling the diameter doubles the radius, and the...
A) p(2p + 1) B) 2p/(2p - 1) C) p D) 1 Solution: The correct answer is B) 2p/(2p - 1) Let v and L stand for the velocity of the sound wave and the length of the organ pipes, respectively. The...
A) [M0 L1 T–3 K–4] B) [M1 L1 T–3 K–3] C) [M1 L2 T–3 K–4] D) [M1 L0 T–3 K–4] Solution: The correct option is D) [M1 L0 T–3 K–4] Expression of power radiated by a body is: P = σAeT4 where σ represents...
A) T/2 B) T/4 C) T/8 D) T/12 Solution: The correct answer is D) T/12 Let's say the particle is in the mean position at time t = 0. As a result, the particle executing S H M has the following...
$ A.\,\,\frac{2M{{R}^{2}}}{3}+\frac{M{{l}^{2}}}{12} $ $ B.\,\,\frac{M{{R}^{2}}}{3}+\frac{M{{l}^{2}}}{12} $ $ C.\,\,\frac{3M{{R}^{2}}}{4}+\frac{M{{l}^{2}}}{12} $ $...
A) 35 m/s B) 70 m/s C) 105 m/s D) 140 m/s Solution: The correct option is According to the question, the speed of sound in air, v...
A) d = h B) d= h/2 C) d = h/4 D) d = 2h Solution: The correct option is D) d = 2h Let R be the earth's radius and g be the acceleration due to gravity at the surface. The following is the...
. The ratio of maximum particle velocity to wave velocity isπa/5 3πa/5 2πa/5 4πa/5 Solution; The correct answer is C. 2πa/5
$ A.\,\,\frac{{{x}^{2}}}{{{A}^{2}}-{{x}^{2}}} $ $ B.\,\,\frac{{{x}^{2}}-{{A}^{2}}}{{{x}^{2}}} $ $ C.\,\,\frac{{{A}^{2}}-{{x}^{2}}}{{{x}^{2}}} $ $ D.\,\,\frac{A-x}{x} $ Solution: The correct option...
A) m2 K2 r2 t2 B) mK2 r2 t C) m K2 r t2 D) m K r2 t Solution: The correct answer is B) mK2 r2 t $ Since\,~{{a}_{c}}={{k}^{2}}r{{t}^{2}} $ $ \frac{{{v}^{2}}}{r}={{k}^{2}}r{{t}^{2}} $ $...
A) 1.4 cm B) 1.8 cm C) 2.0 cm D) 2.2 cm Solution: The correct answer is C) 2.0 cm The equation for capillary action is given as follows: 2S cosθ = ρgrh Or, we can say that: rh=constant r2 h2 = r1 h1...
A) 100Ω B) 200Ω C) 300Ω D) 400Ω Solution: The correct answer is D) 400Ω $ Galvanometer\text{ }resistance=~G $ $ Current\,=\,{{I}_{G}} $ $ Then,\,the\,potential\,difference: $ $ V={{I}_{G}}\left( G+R...
A) the ratio of their amplitudes is 5 B) intensities of individual sources are 25 and 9 units respectively C) the ratio of their amplitudes is 4 D) intensities of individual sources are 4 and 3...
A) 66 B) 67 C) 69 D) 71 Solution: The correct answer is C) 69 $ {{\alpha }_{dc}}=\frac{69}{70} $ $ {{\beta...
henry. If the current in the primary is i = 5 sin (10π t) then the maximum value of e.m.f. induced in coil B isA) π volt B) π/2 volt C) π/3 volt D) π/4 volt Solution: The correct answer is A) π volt $ emf=-M\frac{df}{dt}=-2\times \frac{1}{100}\times \frac{d}{dt}(5sin10\pi t) $ $ emf=-2\times...
A) 2.00 m B) 2.25 m C) 2.50 m D) 2.75 m Solution: The correct answer is D 2.75m According to the question, l=10m Balancing...
A) 0.6 s B) 0.5 s C) 0.4 s D) 0.2 s Solution: The correct option is B) 0.5 s sinθ = 12.5/25 θ = 6π So, the total angle to move is...
A) n B) 2n C) Slightly greater than 2n D) Slightly less than 2n Solution: The correct option is C) Slightly...
$ A.\,\,\frac{YA\alpha L{{t}^{2}}}{2} $ $ B.\,\,\frac{YA{{\alpha }^{2}}L{{t}^{2}}}{2} $ $ C.\,\,\frac{YA{{\alpha }^{2}}{{L}^{2}}{{t}^{2}}}{2} $ $ D.\,\,\frac{YA\alpha Lt}{2} $ Solution: The correct...
A. 59π/900 rad/s B. 59π/1800 rad/s C. 59π/3600 rad/s D. 59π/24000 rad/s Solution: The correct option is B. 59π/1800 rad/s The time taken by minute hand to rotate by an angle of 2π radian is given...
A) force B) impulse C) energy D) momentum Solution: C) energy PV=nRT, where n is the number of moles, is the ideal gas equation. The...
A) demodulation B) modulation C) attenuation D) amplification Solution: The correct option is A) demodulation Modulation is the process of superimposing...
A) P B) Q C) R D) S Solution: The velocity of light in air is c ∴ The velocity of light in a medium, v =...
A) 15 cm to the right B) 15 cm to the left C) 20 cm to the right D) 20 cm to the left Solution: The correct option is A) 15 cm...
A) 3 I and 2 I B) 9 I and 5 I C) 16 I and 3 I D) 25 I and I Solution: The correct option is D) 25 I and I $ Imax={{(\surd I1+\surd...
A) 0.567 eV B) 0.667 eV C) 0.967 eV D) 1.267 eV Solution: The correct option is C) 0.967 eV $ Energy\text{ }of~nth~state\text{ }of\text{...
A) 2μF B) 4μF C) 8μF D) 16μF Solution: The correct option is B) 4μF Let the capacitance of each capacitor be given by C For series...
A) increase in first circuit and decrease in second B) increase in both circuits C) decrease in both circuits D) decrease in first circuit and increase in second Solution: The correct option is D)...
m and 
m. If wavelength of light used is 6000 Å, the fringe formed at that point isA) 10th bright B) 10th dark C) 9th bright D) 9th dark Solution: The correct answer is option(D). The two images of slit act as two new...
A) 0.081 B) 0.162 C) 0.198 D) 0.238 Solution: The correct option is B) 0.162 $ The\,wavelength\text{ }for\text{...
A) small and negative B) small and positive C) large and negative D) large and positive Solution: The correct option is A)...
A) is proportional to mass B) is proportional to impulse C) is inversely proportional to impulse D) does not depend on impulse Solution: The correct option is C) is inversely proportional to impulse...
A) 1 B) 1/2 C) 1/4 D) 1/8 Solution: The correct option is C) 1/4 Initially, the radius of the coil is r1 = r The area of the coil initially is A = πr2 Thus initial magnetic moment of the coil μ1...
A) λ B) 2λ C) 3λ D) 4λ Solution: The correct option is D) 4λ $ Kinetic\text{ }energy\text{ }of\text{ }photoelectrons,~K.E=\frac{hc}{\lambda }-\phi ~ $ $ where\text{ }work\text{ }function,~\phi...
A) 2 : 1 B) 4 : 1 C) 8 : 1 D) 16 : 1 Solution: The correct option is B) 4 : 1 According to the question: C1=4μF, C2=1μF and V=200 volts The equivalent capacitance for series combination is given...
A) 2μF B) 3μF C) 5μF D) 6μF Solution: The correct option is (D) 6μF Expression for the displacement current in the capacitor is, $ i=A{{\varepsilon...
(1) 25 rad/s and 75 rad/s (2) 50 rad/s and 25 rad/s (3) 46 rad/s and 54 rad/s (4) 42 rad/s and 58 rad/s Solution: Answer (3)
(1) 3/4 (2) 7/8 (3) 1/4 (4) 1/8 Solution; Answer (1) M(remain) = 3/4M I = Mremain R2 I = ¾ MR2 So, the correct option is 3/4
(a) R1/R2 (b) R2/R1 (c) R12/R2 (d) R22/R1 Solution: Answer (4) The magnetic field at the centre of the primary coil is given by: $B=\frac{{{\mu }_{0}}I}{2{{R}_{1}}}$ Considering it to be uniform,...
(1) √3Ia2 and 3Ia2 (2) 3 Ia2 and Ia2 (3) 3 Ia2 and Ia2 (4) 4 Ia2 and 3 Ia2 Solution: Answer (1) Current in the loop will be I=V/R which is the same for both loops. Now the magnetic moment of...
(1) 1/2kg (2) 1/3kg (3) 1/6kg (4) 1/12kg Solution: Answer (4) By balancing torque, 2g x 20 = 0.5g x 60 + mg x 120 m = 0.5/6 kg = 1/12kg
Solution: Answer (3)
$ \vec{F}=q(\vec{v}\times \vec{B}) $ $ \vec{F}=q\vec{v}(B\hat{i}+B\hat{j}+{{B}_{0}}\hat{k}) $ $ For\,q=1\,and\,\vec{v}=2\hat{i}+4\hat{j}+6\hat{k}\,and\, $ $ \vec{F}=4\hat{i}-20\hat{j}+12\hat{k} $ $...
(1) 660 V (2) 1320 V (3) 1520 V (4) 1980 V Solution: Answer (4)
(1) 20 cm from the lens, it would be a real image (2) 30 cm from the lens, it would be a real image (3) 30 cm from the plane mirror, it would be a virtual image (4) 20 cm from the plane mirror, it...
(1) 20 m/s, 5 m/s2 (2) 20 m/s, 0 (3) 20 2 m/s, 0 (4) 20 2 m/s, 10 m/s2 Solution: Answer (4) Initial velocity of car = 0 Acceleration of car = 5 m/s2 Velocity of car at t = 4 s; v = u + at Therefore,...
Solution: Answer (4)\ given v = kVe where, k < 1 Thus, v < Ve From conservation of mechanical energy
Solution; Answer (4) T = 2πR/v ⇒ v = 2πR/T
(g = 10 m/s2) nearly (1) 0 kg m/s (2) 4.2 kg m/s (3) 2.1 kg m/s (4) 1.4 kg m/s Solution; Answer (2) According to the question, the mass of the ball =...
(1) 0.2 A (2) 0.4 A (3) 2 A (4) 4 A Solution: Answer (1) In ideal transformer, we know that: Input power = Output power 220 ×...
(1) [F][A][T] (2) [F][A][T2] (3) [F][A][T–1] (4) [F][A–1][T] Solution: Answer (2) We know that, [E] = [Fa][Ab][Tc] => [ML2T–2] = [MLT–2]a [LT–2]b...
(1) 25 (2) 15 (3) 50 (4) 30 Solution: Answer (2) After refraction by a convex lens, parallel beams of light...
(1) [M2] [L–1] [T0] (2) [M] [L–1] [T–1] (3) [M] [L0] [T0] (4) [M2] [L–2] [T–1] Solution: Answer (1) Dimensional formula of energy [E] = [M1 L2 T–2] ...(I) Dimensional formula of gravitational...
$ (1)\,\,{{I}_{d}}={{V}_{0}}\omega C\cos \omega t $ $ (2)\,\,{{I}_{d}}=\frac{{{V}_{0}}}{\omega C}\cos \omega t $ $ (3)\,\,{{I}_{d}}=\frac{{{V}_{0}}}{\omega C}\sin \omega t $ $...
rolls up an inclined plane of angle of inclination 
. The centre of mass of the cylinder has speed of 
. The distance travelled by the cylinder on the inclined surface will be, [take 
] (1) 
(2) 
(3) 
(4) 
Answer (1) SoI. As $v=\sqrt{\frac{2 g h}{1+\frac{K^{2}}{\mathbf{R}^{2}}}}$ $v^{2}=\frac{2 g h}{1+\frac{1}{2}} \quad\left[\frac{K^{2}}{R^{2}}=\frac{1}{2}\right.$ for solid cylinder $]$ $...
, initially at rest, is acted upon by a variable force whose X-component varies with 
in the manner shown. The velocities of the object at the points 
and 
, would have the respective values of (nearly) (1) 
and 
(2) and 
(3) 
and (4) and Answer (4) Sol. Using work-energy theorem $\Delta K=$ work $=$ area under $F=x$ graph From $x=0$ to $x=8 m$ $ \begin{array}{l} \frac{1}{2} m v^{2}=100+30 \\ v^{2}=520 \\ v=\sqrt{520}=23 \mathrm{~m}...
at rest suddenly breaks on its own into three fragments. Two fragments of mass 
each move along mutually perpendicular direction with speed v each. The energy released during the process is, (1) 
(2) 
(3) 
(4) 
Answer (1) Sol. From conservation of linear momentum. $0=m v \hat{j}+m v \hat{i}+3 m v_{1}$ $\vec{v}_{1}=-\frac{v}{3}(\hat{i}+\hat{j})$ $v_{1}=\frac{\sqrt{2}}{3} v$ $K E_{i}=0$ $K E_{f}=\frac{1}{2}...
is kept on a rough horizontal surface (coefficient of friction 
). A horizontal force is applied on the body, but it does not move. The resultant of normal reaction and the frictional force acting on the object is given by F, where 
is, (1) 
(2) 
(3) 
(4) 
Answer (4) Sol. Since body does not move hence it is in equilibrium. $\mathrm{f}_{\mathrm{r}}=$ frictional force which is less than or equal to limiting friction. Now $\mathrm{N}=\mathrm{mg}$ Hence...
and 
. It is observed that the ultimate strength point and the fracture point are close to each other for material but are far apart for material . We can say that materials and are likely to be (respectively), (1) Plastic and ductile (2) Ductile and brittle (3) Brittle and ductile (4) Brittle and plasticAnswer (3) Sol. As given that fracture point and ultimate strength point is close for material $\mathrm{X}$, So, $\mathrm{X}$ is brittle in nature and both points are far apart for material $Y$...
(1) (A) and (B) both are correct. (2) (A) and (B) both are incorrect (3) (A) is correct and (B) is incorrect. (4) (A) is incorrect but (B) is correct. Solution: Answer (3) When reverse biassed, the...
apart and of same height of 
, with the same velocity of 
. When and where will the two bullets collide? 
(1) They will not collide (2) After 
at a height of 
(3) After at a height of 
(4) After 
at a height of 
Answer (2) Sol. Let bullets collide at time t $ \begin{array}{l} x_{1}+x_{2}=100 \mathrm{~m} \\ 25 t+25 t=100 \\ t=2 s \\ \begin{array}{r} y=\frac{1}{2} g t^{2}=\frac{1}{2} \times 10 \times 2^{2} \\...
. The radius of curvature of one of the surfaces is double of the other. Find the radii if the refractive index of the material of the lens is 1.5. (1) 
(2) 
(3) 
(4) 
Answer (4) Sol. Focal length of lens is $\frac{1}{f}=(\mu-1)\left(\frac{1}{R_{1}}-\frac{1}{R_{2}}\right)$, $ \begin{array}{l} \frac{1}{25}=(1.5-1)\left(\frac{1}{R}+\frac{1}{2 R}\right) \\...
Solution : Answer 3 β+ decreases the atomic number by 1, α decreases the atomic number by 2, and the β- decreases the atomic number by 1. Hence the correct order is given by the option 3
(2) 
(3) 
(4) 
Answer (4) Sol. Path difference for destructive interference in YDSE $ \begin{array}{c} \Rightarrow \Delta X_{n}=\frac{(2 n-1)}{2} \lambda \quad n=1,2,3 \ldots . \\ \Delta X_{5^{k+}}=\left(\frac{2...
. It is cut into two symmetrical halves by a plane containing the principal axis. The power of one part will be, (1) (2) 0 (3) 
(4) 
Answer (1) When a lens is cut in half as illustrated in the diagram, the power of each half is the same. i.e. P, because the focal length is the same.
and 
of equal capacitance are connected as shown in the circuit. Terminals a and 
of the key 
are connected to charge capacitor using battery of emf V volt. Now disconnecting a and the terminals and 
are connected. Due to this, what will be the percentage loss of energy? (1) 
(2) 
(3) 
(4) 
Answer (4) $\mathrm{Sol}$, Switch the key at point $c$ $ \begin{array}{l} \frac{q_{0}-q}{C}=\frac{q}{C} \\ 2 q=q_{0} \\ q=\left(\frac{q_{0}}{2}\right) \\...
C. What is the total electric flux across the sphere? (1) 
(2) 
(3) Zero (4) 
Answer (3) Sol. $\phi_{\text {Total }}=\frac{\mathbf{q}_{\text {enclosed }}}{\varepsilon_{0}}$ Because dipoles have equal and opposing charges, there will be no net charge inside the sphere....
(1) Having zero dipole moment (2) Acquire a dipole moment only in the presence of electric field due to displacement of charges (3) Acquire a dipole moment only when magnetic field is absent (4)...
and 
are 
and 
respectively. The velocity of light in this medium would be, (1) 
(2) 
(3) 
(4) 
Answer (2) Sol. $v=\frac{1}{\sqrt{\mu \in}}=\frac{\mathbf{c}}{\sqrt{\mu_{r} \in_{r}}}$ $ \begin{array}{l} =\frac{3 \times 10^{8}}{\sqrt{1 \times 1.44}} \\ =2.5 \times 10^{8} \mathrm{~m} / \mathrm{s}...
is rotated with. constant angular velocity of 10 rad/s in a region of magnetic field of 0.1 T which is perpendicular to the plane of the wheel. The EMF generated between its centre and the rim is, (1) Zero (2) 
(3) 
(4) 
Answer (3) Solution: $ \begin{array}{l} e=\frac{B 1^{2} \omega}{2} \\ =\frac{1}{2} \times 0.1 \times\left(\frac{1}{2}\right)^{2} \times 10 \\ =\frac{1}{8} \\ =0.125 \mathrm{~V} \end{array}...
The capacitance of a parallel plate capacitor is $ C=\frac{{{\in }_{0}}A}{d} $ Potential difference between the plates is V = Ed ........ (ii) Then, the energy stored in the capacitor is given...
gives a current of 
. The same circuit when connected to a DC source of , gives a current of 
. The circuit is (1) Series LCR (2) Series LR (3) Series RC (4) Series LCAnswer (2) Sol. $I_{1}=\frac{V}{Z}$ $ I_{1}=\frac{12}{\sqrt{R^{2}+\left(X_{I}-X_{c}\right)^{2}}}=0.2 \mathrm{~A} $ In the second situation, the capacitor would give infinite resistance, but since...
, at a height 
is radius of earth) from surface of earth. The time period of another satellite whose height is 
from surface will be, (1) 
(2) 
(3) 
(4) 
Answer (2) Sol. $T^{2} \propto r^{3}$ $\mathbf{T}^{2} \propto\left(\mathbf{R}_{\mathrm{E}}+\mathbf{h}\right)^{3}$...
is 
. The number of radioactive atoms in that sample at that instant is, (1) 
(2) 
(3) 
(4) 
Answer (1) Sol. $T_{1 / 2}=\frac{\ln 2}{\lambda}=\frac{0.693}{\lambda}$ $ \begin{array}{l} 2.2 \times 10^{9}=\frac{0.693}{\lambda} \\ \lambda=\frac{0.693}{2.2 \times 10^{9}}=3.15 \times 10^{-10} \\...
suddenly breaks into two pieces whose masses are in the ratio 
. The smaller mass flies off with a velocity 
The velocity of the larger piece will be, (1) 
(2) 
(3) 
(4) 
Answer (2) Sol. By conservation of linear momentum $\overline{\mathbf{P}}=\overrightarrow{\mathbf{P}}_{\mathrm{f}}$ $\mathrm{mv}_{1}=\left(\frac{\mathrm{m}}{6} \overline{\mathrm{v}}_{1}+\frac{5...
(1) 0.9 MeV (2) 9.4 MeV (3) 804 MeV (4) 216 MeV Solution: Answer (4) Reactant mass number = 240 Reactant mass number = 240 The total number of products is 120. BE per product nucleon = 8.5 MeV Total...
and 
. The speed of sound in air is, (1) 
(2) 
(3) 
(4) 
Answer (4) $ \begin{array}{l} I_{1}=9.75 \mathrm{~cm} \\ I_{2}=31.25 \mathrm{~cm} \\ I_{3}=52.75 \mathrm{~cm} \\ e=\text { end correction } \end{array} $ $ \frac{3 \lambda}{4}+e=31.25 \mathrm{~cm} $...