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NEET

Among $[\mathrm{Ni}(\mathrm{CO}) 4],[\mathrm{Ni}(\mathrm{CN}) 4]^{2-}$ and $\left[\mathrm{NiCl}_{4}\right]^{2-}$ species, the hybridisation state of $\mathrm{Ni}$ atoms are respectively:
A $\quad s p^{3}, d s^{2} p, d s p^{2}$

B $\mathrm{sp}^{3}, \mathrm{~d} \mathrm{sp}^{2}, \mathrm{sp}^{3}$
C $\mathrm{sp}^{3}, \mathrm{sp}^{3}, \mathrm{~d} \mathrm{sp}^{2}$
D $\quad$ dsp $^{2}, \mathrm{sp}^{3}, \mathrm{sp}^{3}$

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Correct option is B $\mathrm{sp}^{3}, \mathrm{dsp}^{2}, \mathrm{sp}^{3}$ As CN $^{-}$is a strong ligand so here pairing happens and dsp $^{2}$ hybridisation (square planar) takes place while...

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Q 90. Person with blood group AB is considered as universal recipient because he has: Option A Both A and B antibodies in the plasma. Option B No antigen on the RBC and no antibody in the plasma. Option C Both A and B antigens in the plasma but no antibodies. Option D Both A and B antigens on the RBC but no antibodies in the plasma.

Correct Option: D Solution: A person with blood group AB have both A and B antigens on the RBC but no antibodies in the plasma are considered as universal recipient.

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Q 87. Select the correct option describing gonadotropin activity in a normal pregnant female: Option A High level of FSH and LH facilitate implantation of the embryo. Option B High level of hCG stimulates the synthesis of estrogen and progesterone. Option C High level of hCG stimulates the thickening of endometrium. Option D High level of FSH and LH stimulates the thickening of endometrium

Correct Option: B Solution: In a normal pregnant female high level of hCG (human chorionic gonadotropin) stimulates the synthesis of estrogen and progesterone

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Q 83. Given below is the representation of the extent of global diversity of invertebrates. What groups do the four portions (A-D) represent respectively? A B C D Option A Crustaceans Crustaceans Molluscs Other animal group Option B Molluscs Other animal group Crustaceans Insects Option C Insects Molluscs Crustaceans Other animal group Option D Insects Crustaceans Other animal group Molluscs

Correct Option: C Solution: In the animal kingdom, phylum Arthropoda is the largest with its largest group of insects. The second-largest population is of phylum Mollusca. In the third place is...

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Q 81. Which one of the following is wrongly matched? Option A Translation – Using information in m-RNA to make protein. Option B Repressor protein-Binds to operator to stop enzyme synthesis. Option C Operon – Structural genes, operator and promoter. Option D Transcription – Writing information from DNA to t-RNA

Correct Option: D Solution: Transcription is the process of formation of m-RNA from DNA during protein synthesis. It is not writing information from DNA to t-RNA. Thus this statement is incorrectly...

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Q 72. Dr.F. Went noted that if coleoptile tips were removed and placed on agar for one hour, the agar would produce a bending when placed on one side of freshly cut coleoptile stumps. Of what significance is this experiment? Option A It is the basis for quantitative determination of small amounts of growth promoting substances. Option B It supports the hypothesis that IAA is auxin. Option C It demonstrated polar movement of auxins. Option D It made possible the isolation and exact identification of auxin.

Correct Option: A Solution: The significance of the experiment is that It is the basis for the quantitative determination of small amounts of growth-promoting substances.

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Q 66. How do parasympathetic neural signals affect the working of the heart? Option A Heart rate is increased without affecting the cardiac output. Option B Both heart rate and cardiac output increase. Option C Heart rate decreases but cardiac output increases. Option D Reduces both heart rate and cardiac output.

Correct Option: D Solution: Post-ganglionic fibers of the parasympathetic nervous system secrete acetylcholine. This in return decreases the heart rate and cardiac output.

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Q 64. Geitonogamy involves: Option A Fertilisation of a flower by the pollen from the same flower. Option B Fertilisation of a flower by the pollen from a flower of another plant in the same population. Option C Fertilisation of a flower by the pollen from another flower of plant belonging to a distant population. Option D Fertilisation of a flower by the pollen from another flower of the same plant.

Correct Option: D Solution: Geitonogamy involves fertilization of a flower by the pollen from another flower of the same plant

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Q 59. Fight-or-flight reactions cause activation of: Option A The kidney, leading to suppression of renin angiotensin-aldosterone pathway. Option B The adrenal medulla, leading to increased secretion of epinephrine and norepinephrine. Option C The pancreas, leading to a reduction in the blood sugar levels. Option D The parathyroid glands, leading to an increased metabolic rate.

Correct Option: B Solution: The adrenal medulla, leading to increased secretion of epinephrine and norepinephrine is activated because of the Fight-or-flight reactions. Epinephrine (adrenaline) and...

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Q 48. Match the following and select the correct answer: a. Centriole I Infoldings in mitochondria b. Chlorophyll Ii Thylakoids c. Cristae Iii Nucleic acids d. Ribozymes Iv Basal body cilia or flagella a b c d Option A i. ii. iv. iii. Option B i. iii ii. iv. Option C iv. iii. i. ii. Option D iv. ii. i. iii.

Correct option: D Solution: Centriole is a basal body. Chlorophyll contains thylakoids. Cristae are infoldings present in the inner layer of mitochondria Ribozymes is a catalytic RNA.

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Q 28. At which stage of HIV infection does one usually show symptoms of AIDS? Option A When the infected retro virus enters the host cells. Option B When HIV damages large number of helper T – lymphocytes. Option C When the viral DNA is produced by reverse transcriptase. Option D Within 15 days of sexual contact with an infected person.

Correct Option: B Solution: The helper T-lymphocytes are targeted by the HIV in the blood, causing a decrease in the number of helper T-lymphocytes in the body of the infected person. The infected...

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Q 15. Select the option which is not correct with respect to enzyme action: Option A Addition of a lot of succinate does not reverse the inhibition of succinic dehydrogenase by malonate. Option B A non-competitive inhibitor binds the enzyme at a site distinct from that which binds the substrate. Option C Malonate is a competitive inhibitor of succinic dehydrogenase. Option D Substrate binds with the enzyme at its active site.

Correct option: A Explanation: competitive inhibitor of succinic dehydrogenase is malonate. Malonate competes with the substrate for the substrate-binding site of the enzyme. Similarly, the...

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Q 11. Identify the hormone with its correct matching of source and function. Option A Melatonin – Pineal gland – regulates the normal rhythm of sleep-wake cycle. Option B Progesterone – corpus luteum – secondary sex organs. Option C Atrial natriuretic factor –ventricular wall – increases the blood pressure. Option D Oxytocin – posterior pituitary – growth and maintenance of mammary glands.

Correct option: A Explanation: Melatonin hormone is secreted by the pineal gland, whose function is to regulate the normal rhythm of the sleep-wake cycle.

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Q 6. Just as a person moving from Delhi to Shimla to escape the heat for the duration of hot summer, thousands of migratory birds from Siberia and other extremely cold northern regions move to: Option A Meghalaya Option B Corbett National Park Option C Keoladeo National Park Option D Western Ghat

Correct option: C Explanation: thousands of migratory birds from Siberia and other extremely cold northern regions move to Keoladeo National Park, located in Bharatpur Rajasthan. Presently it is...

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Q 4. Choose the correctly matched pair: Option A Moist surface of buccal cavity – Glandular epithelium Option B Tubular parts of nephrons – Cuboidal epithelium Option C Inner surface of bronchioles – Squamous epithelium Option D Inner lining of salivary ducts – Ciliated epithelium

Correct option: B Explanation: Single layer of cube cells make up the cuboidal epithelium. It is present in the tubular part of the nephron in the kidneys. Its main functions are absorption and...

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In the product

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$ \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} $ $...

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A convex lens ‘A’ of focal length 20 cm and a concave lens ‘B’ of focal length 5 cm are kept along the same axis with a distance ‘d’ between them. If a parallel beam of light falling on ‘A’ leaves ‘B’ as a parallel beam, then the distance ‘d’ in cm will be

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(1) 25                                              (2) 15 (3) 50                                              (4) 30 Solution: Answer (2) After refraction by a convex lens, parallel beams of light...

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A solid cylinder of mass 2 kg and radius $50 \mathrm{~cm}$ rolls up an inclined plane of angle of inclination $30^{\circ}$. The centre of mass of the cylinder has speed of $4 \mathrm{~m} / \mathrm{s}$. The distance travelled by the cylinder on the inclined surface will be, [take $\mathrm{g}=10 \mathrm{~m} / \mathrm{s}^{2}$ ] (1) $2.4 \mathrm{~m}$ (2) $2.2 \mathrm{~m}$ (3) $1.6 \mathrm{~m}$ (4) $1.2 \mathrm{~m}$

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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 $]$ $...

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An object of mass $500 \mathrm{~g}$, initially at rest, is acted upon by a variable force whose X-component varies with $X$ in the manner shown. The velocities of the object at the points $\mathrm{X}=8 \mathrm{~m}$ and $\mathrm{X}=12 \mathrm{~m}$, would have the respective values of (nearly) (1) $18 \mathrm{~m} / \mathrm{s}$ and $20.6 \mathrm{~m} / \mathrm{s}$ (2) $18 \mathrm{~m} / \mathrm{s}$ and $24.4 \mathrm{~m} / \mathrm{s}$ (3) $23 \mathrm{~m} / \mathrm{s}$ and $24.4 \mathrm{~m} / \mathrm{s}$ (4) $23 \mathrm{~m} / \mathrm{s}$ and $20.6 \mathrm{~m} / \mathrm{s}$

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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}...

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A particle of mass $5 \mathrm{~m}$ at rest suddenly breaks on its own into three fragments. Two fragments of mass $m$ each move along mutually perpendicular direction with speed v each. The energy released during the process is, (1) $\frac{4}{3} \mathrm{mv}^{2}$ (2) $\frac{3}{5} \mathrm{mv}^{2}$ (3) $\frac{5}{3} m v^{2}$ (4) $\frac{3}{2} \mathrm{mv}^{2}$

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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}...

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A body of mass $m$ is kept on a rough horizontal surface (coefficient of friction $=\mu$ ). 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 $F$ is, (1) $|\vec{F}|=\mathrm{mg}$ (2) $\overrightarrow{\mid} \mid=m g+\mu m g$ (3) $|\mathrm{F}|=\mu \mathrm{mg}$ (4) $|\vec{F}| \leq m g \sqrt{1+\mu^{2}}$

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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...

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The stress-strain curves are drawn for two different materials $X$ and $Y$. It is observed that the ultimate strength point and the fracture point are close to each other for material $X$ but are far apart for material $Y$. We can say that materials $X$ and $Y$ are likely to be (respectively), (1) Plastic and ductile (2) Ductile and brittle (3) Brittle and ductile (4) Brittle and plastic

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Answer (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 $Y$...

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Two bullets are fired horizontally and simultaneously towards each other from roof tops of two buildings $100 \mathrm{~m}$ apart and of same height of $200 \mathrm{~m}$, with the same velocity of $25 \mathrm{~m} / \mathrm{s}$. When and where will the two bullets collide? $\left(\mathrm{g}=10 \mathrm{~m} / \mathrm{s}^{2}\right)$ (1) They will not collide (2) After $2 \mathrm{~s}$ at a height of $180 \mathrm{~m}$ (3) After $2 \mathrm{~s}$ at a height of $20 \mathrm{~m}$ (4) After $4 \mathrm{~s}$ at a height of $120 \mathrm{~m}$

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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} \\...

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A double convex lens has focal length $25 \mathrm{~cm}$. 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) $50 \mathrm{~cm}, 100 \mathrm{~cm}$ (2) $100 \mathrm{~cm}, 50 \mathrm{~cm}$ (3) $25 \mathrm{~cm}, 50 \mathrm{~cm}$ (4) $18.75 \mathrm{~cm}, 37.5 \mathrm{~cm}$

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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) \\...

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In a Young’s double slit experiment, if there is no initial phase difference between the light from the two slits, a point on the screen corresponding to the fifth minimum has path difference (1) $11 \frac{\lambda}{2}$ (2) $5 \frac{\lambda}{2}$ (3) $10 \frac{\lambda}{2}$ (4) $9 \frac{\lambda}{2}$

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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...

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Two identical capacitors $C_{1}$ and $C_{2}$ of equal capacitance are connected as shown in the circuit. Terminals a and $b$ of the key $k$ are connected to charge capacitor $C_{1}$ using battery of emf V volt. Now disconnecting a and $b$ the terminals $b$ and $c$ are connected. Due to this, what will be the percentage loss of energy? (1) $25 \%$ (2) $75 \%$ (3) $0 \%$ (4) $50 \%$

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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) \\...

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A sphere encloses an electric dipole with charges $\pm 3 \times 10^{-6}$ C. What is the total electric flux across the sphere? (1) $6 \times 10^{-6} \mathrm{Nm}^{2} / \mathrm{C}$ (2) $-3 \times 10^{-6} \mathrm{Nm}^{2} / \mathrm{C}$ (3) Zero (4) $3 \times 10^{-6} \mathrm{Nm}^{2} / \mathrm{C}$

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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....

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For a transparent medium, relative permeability and permittivity, $\mu_{r}$ and $\in_{r}$ are $1.0$ and $1.44$ respectively. The velocity of light in this medium would be, (1) $4.32 \times 10^{8} \mathrm{~m} / \mathrm{s}$ (2) $2.5 \times 10^{8} \mathrm{~m} / \mathrm{s}$ (3) $3 \times 10^{8} \mathrm{~m} / \mathrm{s}$ (4) $2.08 \times 10^{8} \mathrm{~m} / \mathrm{s}$

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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}...

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A cycle wheel of radius $0.5 \mathrm{~m}$ 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) $0.25 \mathrm{~V}$ (3) $0.125 \mathrm{~V}$ (4) $0.5 \mathrm{~V}$

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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}...

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A circuit when connected to an AC source of $12 \mathrm{~V}$ gives a current of $0.2 \mathrm{~A}$. The same circuit when connected to a DC source of $12 \mathrm{~V}$, gives a current of $0.4 \mathrm{~A}$. The circuit is (1) Series LCR (2) Series LR (3) Series RC (4) Series LC

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Answer (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...

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The time period of a geostationary satellite is $24 \mathrm{~h}$, at a height $6 R_{E}\left(R_{E}\right.$ is radius of earth) from surface of earth. The time period of another satellite whose height is $2.5 \mathbf{R}_{E}$ from surface will be, (1) $\frac{12}{2.5} h$ (2) $6 \sqrt{2} \mathrm{~h}$ (3) $12 \sqrt{2} \mathrm{~h}$ (4) $\frac{24}{2.5} \mathrm{~h}$

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Answer (2) Sol. $T^{2} \propto r^{3}$ $\mathbf{T}^{2} \propto\left(\mathbf{R}_{\mathrm{E}}+\mathbf{h}\right)^{3}$...

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The rate of radioactive disintegration at an instant for a radioactive sample of half life $2.2 \times 10^{9} \mathrm{~s}$ is $10^{10} \mathrm{~s}^{-1}$. The number of radioactive atoms in that sample at that instant is, (1) $3.17 \times 10^{19}$ (2) $3.17 \times 10^{20}$ (3) $3.17 \times 10^{17}$ (4) $3.17 \times 10^{18}$

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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} \\...

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An object flying in air with velocity $(20 \hat{i}+25 \hat{j}-12 \hat{k})$ suddenly breaks into two pieces whose masses are in the ratio $1: 5$. The smaller mass flies off with a velocity $(100 \hat{i}+35 \hat{j}+8 \hat{k}) .$ The velocity of the larger piece will be, (1) $-20 \hat{\mathbf{i}}-15 \hat{\mathbf{j}}-80 \hat{k}$ (2) $4 \hat{i}+23 \hat{j}-16 \hat{k}$ (3) $-100 \hat{\mathrm{i}}-35 \hat{\mathrm{j}}-\mathbf{8} \hat{\mathrm{k}}$ (4) $20 \hat{\mathrm{i}}+15 \hat{\mathrm{j}}-80 \hat{\mathrm{k}}$

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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...

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A tuning fork with frequency 800 Hz produces resonance in a resonance column tube with upper end open and lower end closed by water surface. Successive resonance are observed at lengths $9.75 \mathrm{~cm}, 31.25 \mathrm{~cm}$ and $52.75 \mathrm{~cm}$. The speed of sound in air is, (1) $172 \mathrm{~m} / \mathrm{s}$ (2) $500 \mathrm{~m} / \mathrm{s}$ (3) $156 \mathrm{~m} / \mathrm{s}$ (4) $344 \mathrm{~m} / \mathrm{s}$

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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} $...

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A mass falls from a height ‘ $h$ ‘ and its time of fall ‘t’ is recorded in terms of time period $\mathrm{T}$ of a simple pendulum. On the surface of earth it is found that $\mathrm{t}=2 \mathrm{~T}$. The entire set up is taken on the surface of another planet whose mass is half of that of earth and radius the same. Same experiment is repeated and corresponding times noted as ‘ $\mathrm{t}^{\prime}$ and ‘ $\mathrm{T}^{\prime}$. Then we can say (1) $t^{\prime}=2 \mathrm{~T}^{\prime}$ (2) $\mathrm{t}^{\prime}=\sqrt{2} \mathrm{~T}^{\prime}$ (3) $\mathrm{t}^{\prime}>2 \mathrm{~T}^{\prime}$ (4) $\mathrm{t}^{\prime}<2 \mathrm{~T}^{\prime}$

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Answer (1) Sol. For surface of earth time taken in falling $\mathrm{h}$ distance. $ \begin{array}{l} t=\sqrt{\frac{2 h}{g}} \\ \text { and } \quad T=2 \pi \sqrt{\frac{1}{g}} \end{array} $ Given...

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Two metal spheres, one of radius $\mathbf{R}$ and the other of radius $2 R$ respectively have the same surface charge density $\sigma .$ They are brought in contact and separated. What will be the new surface charge densities on them? (1) $\sigma_{1}=\frac{5}{3} \sigma, \sigma_{2}=\frac{5}{6} \sigma$ (2) $\sigma_{1}=\frac{5}{6} \sigma, \sigma_{2}=\frac{5}{2} \sigma$ (3) $\sigma_{1}=\frac{5}{2} \sigma, \sigma_{2}=\frac{5}{6} \sigma$ (4) $\sigma_{1}=\frac{5}{2} \sigma, \sigma_{2}=\frac{5}{3} \sigma$

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Answer (1) Sol. $Q_{1}=\sigma 4 \pi R_{1}{ }^{2}=\sigma 4 \pi R^{2}$ $ \mathbf{Q}_{2}=\sigma 4 \pi(2 R)^{2}=\sigma 16 \pi R^{2} $ After Redistribution of charges $ \begin{array}{l}...

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A straight conductor carrying current i splits into two parts as shown in the figure. The radius of the circular loop is R. The total magnetic field at the centre P of the loop is, (1) $\frac{\mu_{0} i}{2 R}$, inward (2) Zero (3) $3 \mu_{0} / 32 R$, outward (4) $3 \mu_{0} i / 32 R$, inward

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Answer (2) Solution: Net magnetic field at point 'P' $ \mathbf{B}_{\text {net }}=\overrightarrow{\mathbf{B}_{1}}+\overline{\mathbf{B}_{2}} $ Here $\bar{B}_{1}$ and $\bar{B}_{2}$ are equal in...

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A truck is stationary and has a bob suspended by a light string, in a frame attached to the truck. The truck suddenly moves to the right with an acceleration of a. The pendulum will tilt (1) to the left and angle of inclination of the pendulum with the vertical is $\tan ^{-1}\left(\frac{\mathbf{g}}{\mathbf{a}}\right)$ (2) to the left and angle of inclination of the pendulum with the vertical is $\sin ^{-1}(\underline{\mathbf{g}}{\mathrm{a}})$(3) to the left and angle of inclination of tha pendulum with the vertical is $\tan ^{-1}\left(\frac{\mathbf{a}}{\mathbf{g}}\right)$ (4) to the left and angle of inclination of the pendulum with the vertical is $\sin ^{-1}\left(\frac{\mathbf{a}}{\mathbf{g}}\right)$

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Answer (3) Sol. From the frame of truck $ \begin{array}{l} \operatorname{Tsin} \theta=\mathbf{m a} \\ T \cos \theta=\mathrm{mg} \end{array} $ $\theta=\tan...

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A person standing on drops a coin. The coin reaches the floor in time $\mathrm{t}_{1}$ if the elevator is at rest and in time $\mathrm{t}_{2}$ if the elevator is moving uniformly. Then (1) $t_{1}=t_{2}$ (2) $t_{1}t_{2}$ depending upon whether the lift is going up or down (3) $\mathrm{t}_{1}<\mathrm{t}_{2}$ (4) $t_{4}>t_{2}$

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Answer (1) Sol. Because the elevator is an inertial frame of reference in both cases, the effective gravity is the same in both cases. So, time of fall remains same in both cases (since initial...

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A particle starting from rest, moves in a circle of radius ‘ $r^{\prime} .$ It attains a velocity of $\mathrm{V}_{0} \mathrm{~m} / \mathrm{s}$ in the $n^{\text {th }}$ round. Its angular acceleration will be, (1) $\frac{\mathrm{V}_{0}^{2}}{4 \pi \mathrm{nr}} \mathrm{rad} / \mathrm{s}^{2}$ (2) $\frac{\mathrm{V}_{0}}{\mathrm{n}} \mathrm{rad} / \mathrm{s}^{2}$ (3) $\frac{\mathrm{V}_{0}^{2}}{2 \pi \mathrm{nr}^{2}} \mathrm{rad} / \mathrm{s}^{2}$ (4) $\frac{\mathrm{V}_{0}^{2}}{4 \pi \mathrm{rr}^{2}} \mathrm{rad} / \mathrm{s}^{2}$

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Answer (4) Sol. Initial speed $=0$ Final speed = Tangential acceleration=$\alpha r$ $V^{2}=u^{2}+2 a s$ $V_{0}^{2}=0+2 r \alpha(2 \pi r) n$ $\alpha=\frac{V_{0}^{2}}{4 \pi n r^{2}}$

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Two small spherical metal balls, having equal masses, are made from materials of densities $\rho_{1}$ and $\rho_{2}\left(\rho_{1}=8 \rho_{2}\right)$ and have radii of $1 \mathrm{~mm}$ and $2 \mathrm{~mm}$, respectively, they are made to fall vertically (from rest) in a viscous medium whose coefficient of viscosity equals $\eta$ and whose density is $0.1 \rho_{2} .$ The ratio of their terminal velocities would be, (1) $\frac{79}{36}$ (2) $\frac{79}{72}$ (3) $\frac{19}{36}$ (4) $\frac{39}{72}$

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Answer (1) Solution: As $V_{T}=\frac{2 a^{2}}{9 \eta}(\rho-\sigma) g$ $ \begin{array}{l} V_{T_{1}}=\frac{2 \times(1)^{2}}{9 \eta}\left(\rho_{1}-0.1 \rho_{2}\right) g \\ V_{T_{1}}=\frac{2 \times 1}{9...

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Angular width of the central maxima in the Fraunhofer diffraction for $\lambda=6000 \AA$ is $\theta_{0^{+}}$ When the same slit is illuminated by another monochromatic light, the angular width decreases by $30 \%$. The wavelength of this light is (1) $420 \mathrm{~A}$ (2) $1800 \mathrm{~A}$ (3) $4200 \mathrm{~A}$ (4) $6000 \AA$

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Answer (3) Sol. As $\theta=\frac{2 \lambda}{a}$ $ \begin{array}{l} \theta_{0}=\frac{2 \times 6000}{\mathbf{a}} \\ \frac{\theta^{\prime}}{\theta_{0}}=\frac{\lambda^{\prime}}{6000} \\ \Rightarrow...

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$1 \mathrm{~g}$ of water, of volume $1 \mathrm{~cm}^{3}$ at $100^{\circ} \mathrm{C}$, is converted into steam at same temperature under normal atmospheric pressure $\left(\therefore 1 \times 10^{5} \mathrm{~Pa}\right)$. The volume of steam formed equals $1671 \mathrm{~cm}^{3}$. If the specific latent heat of vaporisation of water is $2256 \mathrm{~J} / \mathrm{g}$, the change in internal energy is, (1) $2256 \mathrm{~J}$ (2) $2423 \mathrm{~J}$ (3) $2089 \mathrm{~J}$ (4) $167 \mathrm{~J}$

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Answer (3) Sol. $ \begin{aligned} \Delta Q &=2256 \times 1=2256 \mathrm{~J} \\ \Delta \mathrm{W} &=P\left[V_{\text {steam }}-V_{\text {water }}\right] \\ &=1 \times 10^{5}[1671-1] \times...

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Assuming that the gravitational potential energy of an object at infinity is zero, the change in potential energy (final – initial) of an object of mass $\mathbf{m}$, when taken to a height $\mathbf{h}$ from the surface of earth (of radius $\mathbf{R}$ ), is given by, (1) $\frac{\mathrm{GMm}}{\mathbf{R}+\mathrm{h}}$ (2) $=\frac{\mathrm{GMm}}{\mathrm{R}+\mathrm{h}}$ (3) $\frac{\mathrm{GMmh}}{\mathrm{R}(\mathrm{R}+\mathrm{h})}$ (4) $\mathrm{mgh}$.

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Answer (3) $(\mathrm{P} . \mathrm{E})_{\mathrm{A}}=-\frac{\mathrm{GMm}}{\mathrm{R}}$ $(P . E)_{B}=-\frac{G M m}{R+h}$ $\therefore \Delta U=(P . E)_{B}-(P . E)_{A}$ $=-\frac{G M m}{R+h}+\frac{G M...

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A person travelling in a straight line moves with a constant velocity $v_{1}$ for certain distance ${ }^{\prime} \mathrm{x}^{3}$ and with a constant velocity $v_{2}$ for next equal distance. The average velocity $v$ is given by the relation (1) $v=\sqrt{v_{1} v_{2}}$ (2) $\frac{1}{v}=\frac{1}{v_{1}}+\frac{1}{v_{2}}$ (3) $\frac{2}{v}=\frac{1}{v_{1}}+\frac{1}{v_{2}}$ (4) $\frac{v}{2}=\frac{v_{1}+v_{2}}{2}$

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Answer (3) Sol. As $t_{1}=\frac{x}{v_{1}}$ and $t_{2}=\frac{x}{v_{2}}$ $ \begin{array}{l} \therefore \quad v=\frac{x+x}{t_{1}+t_{2}} \\ \qquad=\frac{2 x}{\frac{x}{v_{1}}+\frac{x}{v_{2}}}=\frac{2...

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The main scale of a vernier callipers has $n$ divisions/cm. $n$ divisions of the vernier scale coincide with $(\mathrm{n}-1)$ divisions of main scale. The least count of the vernier callipers is, (1) $\frac{1}{n(n+1)} \mathrm{cm}$ (2) $\frac{1}{(n+1)(n-1)} c m$ (3) $\frac{1}{n} \mathrm{~cm}$ (4) $\frac{1}{n^{2}} \mathrm{~cm}$

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Answer (4) Sol. $n V S D=(n-1)$ MSD $1 \mathrm{VSD}=\frac{(\mathrm{n}-1)}{\mathrm{n}} \mathrm{MSD}$ $ \begin{aligned} \text { L.C. } &=1 \mathrm{MSD}-1 \mathrm{VSD} \\ &=1...

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The value of $\gamma\left(=\frac{\mathbf{C}_{p}}{\mathrm{C}_{\mathrm{v}}}\right)$, for hydrogen, helium and another ideal diatomic gas $X$ (whose molecules are not rigid but have an additional vibrational mode), are respectively equal to, (1) $\frac{7}{5}, \frac{5}{3}, \frac{7}{5}$ (2) $\frac{7}{5}, \frac{5}{3}, \frac{9}{7}$ (3) $\frac{5}{3}, \frac{7}{5}, \frac{9}{7}$ (4) $\frac{5}{3}, \frac{7}{5}, \frac{7}{5}$

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Answer (2) Sol. If the vibrational mode is ignored, diatomic gases have 5 degrees of freedom. When the vibrational mode is taken into account, the degrees of freedom of diatomic gas molecules...

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A deep rectangular pond of surface area $A$, containing water (density $=\rho$ ), specific heat capacity $=\mathrm{s}$ ), is located in a region where the outside air temperature is at a steady value of $-26^{\circ} \mathrm{C}$. The thickness of the frozen ice layer in this pond, at a certain instant is $\mathrm{x}$. Taking the thermal conductivity of ice as $K$, and its specific latent heat of fusion as $L$, the rate of increase of the thickness of ice layer, at this instant, would be given by (1) $26 \mathrm{~K} / \rho \times(\mathrm{L}+4 \mathrm{~s})$ (2) $26 \mathrm{~K} / \rho \times(\mathrm{L}-4 \mathrm{~s})$ (3) $26 K /\left(\rho x^{2} L\right)$ (4) $26 \mathrm{~K} /(\rho \times \mathrm{L})$

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Answer (4) Sol. Assume at any instant thickness of ice is $\mathrm{x}$. And time taken to form additional thickness. $(\mathrm{dx})$ is $\mathrm{dt}$. $\mathrm{mL}=\frac{\mathrm{KA}[26-0] \mathrm{d}...

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In a u-tube as shown in the fig. water and oil are in the left side and right side of the tube respectively. The heights from the bottom for water and oil columns are $15 \mathrm{~cm}$ and $20 \mathrm{~cm}$ respectively. The density of the oil is [take $\left.\rho_{\text {water }}=1000 \mathrm{~kg} / \mathrm{m}^{3}\right]$ (1) $1333 \mathrm{~kg} / \mathrm{m}^{3}$ (2) $1200 \mathrm{~kg} / \mathrm{m}^{3}$ (3) $750 \mathrm{~kg} / \mathrm{m}^{3}$ (4) $1000 \mathrm{~kg} / \mathrm{m}^{3}$

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  Answer (3) Sol. In equilibrium Pressure at $A=$ Pressure at $B$. $ P_{a}+0.15 \times 10^{3} \times g=P_{a}+0.20 \times d_{0} g $ $ \begin{aligned} d_{0} &=\frac{0.15 \times 10^{3}}{0.20}...

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The relations amongst the three elements of earth’s magnetic field, namely horizontal component $\mathrm{H}$, vertical component $\mathrm{V}$ and $\operatorname{dip} 8$ are, $\left(B_{E}=\right.$ total magnetic field) (1) $\mathrm{V}=\mathbf{B}_{\mathrm{E}}, \mathrm{H}=\mathrm{B}_{\mathrm{E}} \tan \delta$ (2) $V=B_{E}$ tan\delta, $H=B_{E}$ (3) $\mathrm{V}=\mathbf{B}_{\mathrm{E}} \sin \delta, \mathrm{H}=\mathrm{B}_{\mathrm{E}} \cos \delta$ (4) $V=B_{E} \cos \delta, H=B_{E} \sin \delta$

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Answer (3) Sol. $H=B_{E} \cos \delta_{1}$ $ \mathrm{V}=\mathbf{B}_{\mathrm{E}} \sin \delta^{\mathrm{i}} $

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The metre bridge shown is in balance position with $\frac{P}{Q}=\frac{L_{1}}{l_{2}}$. If we now interchange the positions of galvanometer and cell, will the bridge work? If yes, what will be balance condition? (1) yes, $\frac{P}{Q}=\frac{I_{1}}{l_{2}}$ (2) yes, $\frac{P}{Q}=\frac{l_{2}-L_{1}}{l_{2}+l_{1}}$ (3) no, no null point $(4)$ yes, $\frac{P}{Q}=\frac{I_{2}}{I_{1}}$

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Answer (1) Sol. In balanced bridge (initially) In balanced bridge (finally) $ \begin{array}{l} \frac{P}{L_{1}}=\frac{Q}{l_{2}} \\ \frac{P}{Q}=\frac{L_{1}}{l_{2}} \end{array} $ The equilibrium state...

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 In a potentiometer circuit a cell of EMF 1.5 V gives balance point at 36 cm length of wire. If another cell of EMF 2.5 V replaces the first cell, then at what length of the wire, the balance point occurs?

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(1) 60 cm                                         (2) 21.6 cm (3) 64 cm                                         (4) 62 cm Solution: Answer (1) From the application of potentiometer to compare two...

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The radius of the first permitted Bohr orbit, for the electron, in a hydrogen atom equals $0.51 \AA$ and its ground state energy equals $-13.6 \mathrm{eV}$. If the electron in the hydrogen atom is replaced by muon $\left(\mu^{-}\right)$[charge same as electron and mass $\left.207 \mathrm{~m}_{\mathrm{e}}\right]$, the first Bohr radius and ground state energy will be, (1) $2.56 \times 10^{-13} \mathrm{~m},-13.6 \mathrm{eV}$ (2) $0.53 \times 10^{-13} \mathrm{~m},-3.6 \mathrm{eV}$ (3) $25.6 \times 10^{-13} \mathrm{~m},-2.8 \mathrm{eV}$ (4) $2.56 \times 10^{-13} \mathrm{~m},-2.8 \mathrm{keV}$

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The correct answer is (4) Sol. $r_{n} \propto \frac{1}{m}$ $ \begin{array}{l} \mathrm{r}_{\mu}=\frac{0.51}{207}=2.56 \times 10^{-13} \mathrm{~m} \\ \mathrm{E} \propto \mathrm{m}_{\mathrm{e}} \\...

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A beam of light consisting of red, green and blue colours is incident on a right angled prism. The refractive index of the material of the prism for the above red, green and blue wavelengths are $1.39,1.44$ and $1.47$ respectively. The prism will: (1) Separate the red colour part from the green and blue colours (2) Separate the blue colour part from the red and green colours (3) Separate all the three colours from one another. (4) Not separate the three colours at all

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Solution: (1) For $\mathrm{TIR} I>\mathrm{I}_{\mathrm{c}}$ so $\operatorname{Sin} \mathrm{i}>\sin \mathrm{I}_{\mathrm{c}}$ $ \operatorname{Sin} 45^{\circ}>\frac{1}{\mu} \Rightarrow \mu...

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A photoelectric surface is illuminated successively by monochromatic light of wavelength $\lambda$ and $\frac{\lambda}{2}$. If the maximum kinetic energy of the emitted photoelectrons in the second case is 3 times that in the first case, the work function of the surface of the material is: (h $=$ Planck’s constant, $c=$ speed of light) (1) $\frac{h c}{3 \lambda}$ (2) $\frac{h c}{2 \lambda}$ (3) $\frac{h c}{\lambda}$ (4) $\frac{2 h c}{\lambda}$

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Solution: (2) $ \begin{array}{l} k_{1}=\frac{h c}{\lambda}-\psi \\ k_{2}=3 k_{1}=\frac{2 h c}{\lambda}-\psi=\frac{3 h c}{\lambda}-3 \psi \end{array} $ So $2 \psi=\frac{h c}{\lambda}$ So...

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Water rises to a height ‘ $h$ ‘ in capillary tube. If the length of capillary tube above the surface of water is made less than ‘ $\mathrm{h}$ ‘, then: (1) Water does not rise at all (2) Water rises upto the tip of capillary tube and then starts overflowing like a fountain (3) Water rises upto the top of capillary tube and stays there without overflowing (4) Water rises upto a point a little below the top and stays there

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Solution: (3) Water will not overflow, but its radius of curvature will vary. Explanation: It won't overflow. That should be obvious since doing so would create a constant flow, constantly using...

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If dimensions of critical velocity $v_{c}$ of a liquid flowing through a tube are expressed as $\left[\eta^{x} \rho^{y} r^{2}\right]$, where $\eta, \rho$ and $r$ are the coefficient of viscosity of liquid, density of liquid and radius of the tube respectively, then the values of $x, y$ and $z$ are given by: (1) $\quad 1,1,1$ (2) $1,-1,-1$ (3) $-1,-1,1$ (4) $\quad-1,-1,-1$

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Solution: (2) $ \mathrm{V}_{\mathrm{c}}=\eta^{\mathrm{x}} \rho^{\mathrm{y}} \mathrm{r}^{\mathrm{z}} $ Critical velocity is given by $V_{c}=\frac{R \eta}{2 \rho r}$ So, $x=1$ $ \begin{aligned} y=-1...

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The input signal given to a CE amplifier having a voltage gain of 150 is $\mathrm{V}_{\mathrm{i}}=2 \cos \left(15 \mathrm{t}+\frac{\pi}{3}\right)$. The corresponding output signal will be: (1) $300 \cos \left(15 \mathrm{t}+\frac{4 \pi}{3}\right)$ (2) $\quad 300 \cos \left(15 \mathrm{t}+\frac{\pi}{3}\right)$ (3) $75 \cos \left(15 \mathrm{t}+\frac{2 \pi}{3}\right)$ (4) $2 \cos \left(15 t+\frac{5 \pi}{6}\right)$

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Solution: (1) The phase difference caused by CE amplifier $\pi\left(=180^{\circ}\right)$ so $V_{\text {out }}=300 \cos \left(15 t+\frac{\pi}{3}+\pi\right)$

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A proton and an alpha particle both enter a region of uniform magnetic field B, moving at right angles to the field B. if the radius of circular orbits for both the particles is equal and the kinetic energy acquired by proton is $1 \mathrm{MeV}$, the energy acquired by the alpha particle will be: (1) $1 \mathrm{MeV}$ (2) $4 \mathrm{MeV}$ (3) $0.5 \mathrm{MeV}$ (4) $1.5 \mathrm{MeV}$

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Solution: (1) $ \mathrm{R}=\frac{\mathrm{mV}}{\mathrm{qB}}=\frac{\sqrt{2 \mathrm{~m}(\mathrm{kE})}}{\mathrm{qB}} $ Since $\mathrm{R}$ is same so $\mathrm{KE} \propto...

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The heart of a man pumps 5 litres of blood through the arteries per minute at a pressure of $150 \mathrm{~mm}$ of mercury. If the density of mercury be $13.6 \times 10^{3} \mathrm{~kg} / \mathrm{m}^{3}$ and $\mathrm{g}=10 \mathrm{~m} / \mathrm{s}^{2}$ then the power of heart in watt is: (1) $1.50$ (2) $1.70$ (3) $\quad 2.35$ (4) $3.0$

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Solution: (2) $ \begin{aligned} \text { Power } &=\vec{F}, \vec{V}=P A \vec{V}=\rho g h A V \\ &=13.6 \times 10^{3} \times 10 \times 150 \times 10^{-3} \times 0.5 \times 10^{-3} /_{60} \text...

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If potential (in volts) in a region is expressed as $V(x, y, z)=6 x y-y+2 y z$, the electric field (in $N / C)$ at point $(1,1,0)$ is: (1) $-(6 \hat{1}+9 \hat{\jmath}+\hat{k})$ (2) $\quad-(3 \hat{1}+5 \hat{j}+3 \hat{k})$ (3) $-(6 \hat{i}+5 \hat{\jmath}+2 \hat{k})$ (4) $-(2 \hat{1}+3 \hat{j}+\mathrm{k})$

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Solution: (3) $ \begin{array}{l} V=6 x y-y+24 z \\ \bar{E}=\left(\frac{\partial V}{\partial x} \hat{1}+\frac{\partial V}{\partial y} \hat{j} \frac{\partial V}{\partial z} \hat{k}\right) \\...

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On a frictionless surfaces, a block of mass $M$ moving at speed $v$ collides elastically with another block of same mass $M$ which is initially at rest. After collision the first block moves at an angle $\theta$ to its initial direction and has a speed $\frac{\mathrm{v}}{3}$. The second block’s speed after the collision is: (1) $\frac{\sqrt{3}}{2} v$ (2) $\frac{2 \sqrt{2}}{3} v$ (3) $\frac{3}{4} v$ (4) $\frac{3}{\sqrt{2}} v$

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Solution: (2) $\overrightarrow{\mathrm{P}}_{1}=\overrightarrow{\mathrm{P}_{\mathrm{f}}}$ $\Rightarrow\left|\mathrm{P}_{\mathrm{i}}\right|=\left|\mathrm{P}_{\mathrm{f}}\right| \Rightarrow...

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A ball is thrown vertically downwards from a height of $20 \mathrm{~m}$ with an initial velocity $\mathrm{V}_{0}$. It collides with the ground, loses 50 percent of its energy in collision and rebounds to the same height. The initial velocity $v_{0}$ is: (Take $g=10 \mathrm{~ms}^{-2}$ ) (1) $10 \mathrm{~ms}^{-1}$ (2) $14 \mathrm{~ms}^{-1}$ (3) $20 \mathrm{~ms}^{-1}$ (4) $28 \mathrm{~ms}^{-1}$

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Solution: (3) $ \begin{array}{l} \frac{K E_{f}}{\mathbb{K E}_{i}}=\frac{1}{2} \\ \frac{V_{f}}{V_{i}}=\frac{1}{\sqrt{2}} \\ \frac{\sqrt{2 g h}}{\sqrt{V_{0}^{2}+2 g h}}=\frac{1}{\sqrt{2}} \end{array}...

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Two stones of masses ma and $2 \mathrm{~m}$ are whirled in horizontal circles, the heavier one in a radius $\frac{\mathrm{r}}{2}$ and the lighter one in radius r. the tangential speed of lighter stone is $n$ times that of the value of heavier stone when they experience same centripetal forces. The value of $n$ is: (1) 1 (2) 2 (3) 3 (4) 4

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The Solution is (2) $ \mathrm{F}_{\mathrm{C}}=\frac{\mathrm{mv}_{1}^{2}}{\mathrm{r}}=\frac{2 \mathrm{mv}_{2}^{2}}{\left(\frac{\mathrm{r}}{2}\right)}=\frac{4 \mathrm{mv}_{2}^{2}}{\mathrm{r}} $...

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A plank with a box on it at one end is gradually raised about the other end. As the angle of inclination with the horizontal reaches $30^{\circ}$, the box starts to slip and slides $4.0 \mathrm{~m}$ down the plank in $4.0 \mathrm{~s}$. The coefficients of static and kinetic friction between the box and the plank will be, respectively: (1) $0.4$ and $0.3$ (2) $0.6$ and $0.6$ (3) $0.6$ and $0.5$ (4) $0.5$ and $0.6$

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The Solution is (3) $ \begin{array}{l} \mu_{s}=\tan 30^{\circ}=\frac{1}{\sqrt{3}}=0.5 \\ \mu_{s}=0.57=0.6 \\ \mathrm{~S}=\mathrm{ut}+\frac{1}{2} \mathrm{at}^{2} \\ 4=\frac{1}{2} a(4)^{2} \Rightarrow...

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A screw gauge gives the following readings when used to measure the diameter of a wire Main scale reading : 0 mm Circular scale reading : 52 divisions Given that 1 mm on main scale corresponds to 100 divisions on the circular scale. The diameter of the wire from the above data is

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(1) 0.52 cm                           (2) 0.026 cm (3) 0.26 cm                           (4) 0.052 cm Solution: Answer (4) According to the question, the pitch of the screw gauge, P = 1 mm The...

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A series R-C circuit is connected to an alternating voltage source. Consider two situations: i. When capacitor is air filled. ii. When capacitor is mica filled. Current through resistor is $i$ and voltage across capacitor is $V$ then: (1) $V_{a}=V_{b}$ (2) $V_{a}\mathrm{V}_{\mathrm{b}}$ (4) $\quad \mathrm{l}_{\mathrm{a}}>\mathrm{i}_{\mathrm{b}}$

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Solution: (3) $ \begin{array}{l} X_{c}=1 / \mathrm{C} \mathrm{N} \end{array} $ $ \begin{array}{l} i=\frac{v}{\sqrt{R^{2}+\left(\frac{1}{c \omega}\right)^{2}}} \\...

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