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Physics NEET Questions — Free MCQ Practice

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A planet moving along an elliptical orbit is closest to the sun at a distance $r_{1}$ and farthest away at a distance of $r_{2}$ . If $v_{1} and v_{2}$ are the linear velocities at these points respectively, then the ratio $\frac{v_{1}}{v_{2}}$ is

A.

$r_{2} / r_{1}$

B.

$(r_{2} / r_{1})^{2}$

C.

$r_{1} / r_{2}$

D.

$(r_{1} / r_{2})^{2}$

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Explanation

From the law of conservation of angular momentum

${mr}_{1} v_{1} = {mr}_{2} v_{2} r_{1} v_{1} = r_{2} v_{2} \frac{v_{1}}{v_{2}} = \frac{r_{2}}{r_{1}}$

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A body projected vertically from the earth reaches a high equal to earth's radius before returning to the earth. The power exerted by the gravitational force is greatest

A.

at the instant just before the body hits the earth

B.

it remains constant all through

C.

at the instant just after the body is projected

D.

at the highest position of the body

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Explanation

We know

$p = F \cdot v = F v \cos θ$

So just before hitting $θ$ is zero and both F and $v$ are maximum.

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A particle of mass m is thrown upwards from the surface of the earth with a velocity u. The mass and the radius of the earth are, respectively, M and R. G is gravitational constant and g is acceleration due to gravity on the surface of the earth. The minimum value of u so that the particle does not return back to earth is

A.

$\sqrt{\frac{2 G M}{R}}$

B.

$\sqrt{\frac{2 G M}{R^{2}}}$

C.

$\sqrt{2 g R^{2}}$

D.

$\sqrt{\frac{2 G M}{R^{2}}}$

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Explanation

Escape velocity

$V_{e} = \sqrt{2 g R} V_{e} = \sqrt{\frac{2 G M}{R^{2}} R} (∵ G M = g R^{2}) V_{e} = \sqrt{\frac{2 G M}{R}}$

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The radii of circular orbits of two satellites A and B of the earth are 4R and R, respectively.If the speed of satellite A is 3v, then speed of satellite B will be

A.

3v/4

B.

6v

C.

12v

D.

3v/2

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Explanation

Orbital velocity of satellite v= $\sqrt{\frac{G M}{r}}$

$\Rightarrow \frac{v_{A}}{v_{B}} = \sqrt{\frac{r_{B}}{r_{A}}} = \sqrt{\frac{R}{4 R}} = \frac{1}{2}$

$∴ \frac{v_{A}}{v_{B}} = \frac{3 v}{v_{B}} = \frac{1}{2}$

$∴ v_{B} = 6 v$

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The additional kinetic energy to be provided to a satellite of mass m revolving around a planet of mass M, to transfer it from a circular orbit of radius $R_{1}$ to another of radius $R_{2} (R_{2} > R_{1})$ is

A.

$G m M (\frac{1}{R_{1}^{2}} - \frac{1}{R_{2}^{2}})$

B.

$G m M (\frac{1}{R_{1}} - \frac{1}{R_{2}})$

C.

$2 G m M (\frac{1}{R_{1}} - \frac{1}{R_{2}})$

D.

$\frac{1}{2} G m M (\frac{1}{R_{1}} - \frac{1}{R_{2}})$

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Explanation

$Total KE required to change the orbit will be equal to change in total energy as PE as well as KE, both are changing . ∆ E = (- \frac{G M m}{2 R_{2}}) - (- \frac{G M m}{2 R_{1}}) {KE}_{extra} = \frac{G M m}{2} [\frac{1}{R_{1}} - \frac{1}{R_{2}}]$

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A gas undergoes a process in which its pressure P and volume V are related as ${VP}^{n} = constant$ . The bulk modulus for the gas in the process is
[Only for Dropper and XII batch]

A.

$\frac{P}{n}$

B.

$P^{1 / n}$

C.

nP

D.

$P^{n}$

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Explanation

$(3) D i f f e r e n t i a t e V P^{n} = C w . r . t o V u \sin g m u l t i p l i c a t i o n r u l e : V \frac{d}{d V} (P^{n}) + P^{n} \frac{d}{d V} (V) = 0 V n P^{n - 1} \frac{d P}{d V} + P^{n} . 1 = 0 V n P^{n - 1} d P + P^{n} . d V = 0 n V P^{n - 1} d P = - P^{n} . d V \Rightarrow \frac{d P}{d V} = - \frac{P^{n}}{n V P^{n - 1}} = - \frac{P}{n V} B u l k M o d u l u s = - \frac{d P}{\frac{d V}{V}} = - (- \frac{P}{\frac{n V}{V}}) = \frac{P}{n}$

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The breaking stress of a wire depends upon

A.

material of the wire

B.

length of the wire

C.

radius of the wire

D.

shape of the cross section

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Explanation

Breaking stress depends only on nature of material

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The elastic energy stored in a wire of Young's Modulus Y is -

A.

$Y \times \frac{{strain}^{2}}{volume}$

B.

$stress \times strain \times volume$

C.

$\frac{{strain}^{2} \times volume}{2 Y}$

D.

$\frac{1}{2} \times s t r e s s \times s t r a i n \times v o l u m e$

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Explanation

$U = \frac{1}{2} F ∆ l = \frac{1}{2} σ A ∆ l = \frac{1}{2} σ \frac{∆ l}{l} A l = \frac{1}{2} σ ε V = \frac{1}{2} \times s t r e s s \times s t r a i n \times v o l u m e$

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If Young modulus (Y) equal to bulk modulus (B). Then the Poisson ratio is :

A.

$\frac{1}{3}$

B.

$\frac{2}{3}$

C.

$\frac{1}{2}$

D.

$\frac{1}{4}$

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Explanation

$Y = 3 B (1 - 2 σ)$

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The bulk modulus of a spherical object is B. If it is subjected to uniform pressure p, the fractional decrease in radius is

A.

$\frac{P}{B}$

B.

$\frac{B}{3 p}$

C.

$\frac{3 p}{B}$

D.

$\frac{p}{3 B}$

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Explanation

(d)The object is spherical and the bulk modulus is represented by B. It is the ratio of normal stress to the volumetric strain.

Hence $B = \frac{F / A}{∆ V / V}$

$\Rightarrow \frac{∆ V}{V} = \frac{P}{B} \Rightarrow |\frac{∆ V}{V}| = \frac{P}{B}$

Hence p is applied pressure on the object and $\frac{∆ V}{V}$ is

volume strain

Fractional decrease in volume

$\Rightarrow \frac{∆ V}{V} = 3 \frac{∆ R}{R} [S o, V = \frac{4}{3} {πR}^{3}]$

Volume of the sphere decreases due to the decrease in its radius.

Hence $\frac{∆ V}{V} = \frac{3 ∆ R}{R} = \frac{P}{B} \Rightarrow \frac{∆ R}{R} = \frac{P}{3 B}$

Physics MCQs for NEET — Practice Questions with Answers

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