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NEET NEET Questions — Free MCQ Practice

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A long spring is stretched by 2 cm, its potential energy is U. If the spring is streched by 10 cm, find the potential energy stored in it. [This question is only for Dropper and XII batch]

A.

10 U

B.

15 U

C.

20 U

D.

25 U

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Explanation

Elastic potential energy of a spring $U = \frac{1}{2} k x^{2} ∴ U \propto x^{2}$

So $\frac{U_{2}}{U_{1}} = {(\frac{x_{2}}{x_{1}})}^{2} \Rightarrow \frac{U_{2}}{U} = {(\frac{10 c m}{2 c m})}^{2} \Rightarrow U_{2} = 25 U$

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A spring of spring constant $5 \times 10^{3} N / m$ is stretched initially by 5 cm from the unstretched position. Find the work required to stretch it further by another 5 cm is [This question is only for Dropper and XII batch]

A.

15 J

B.

18.75 N .m

C.

20 J

D.

22.75 N .m

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Explanation

Work done to stretch the spring from $x_{1} t o x_{2}$

$W = \frac{1}{2} k (x_{2}^{2} - x_{1}^{2}) = \frac{1}{2} 5 \times 10^{3} [{(10 \times 10^{- 2})}^{2} - {(5 \times 10^{- 2})}^{2}] = \frac{1}{2} \times 5 \times 10^{3} \times 75 \times 10^{- 4} = 18.75 N . m .$

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An automobile of mass m accelerates, starting from rest, while the engine supplies constant power P, its position and velocity changes w.r.t time as- [This question is only for Dropper and XII batch]

A.

${(\frac{P t}{9 m})}^{\frac{1}{2}} t^{1 / 2}, {(\frac{2 P t}{m})}^{\frac{1}{2}}$

B.

${(\frac{2 P t}{9 m})}^{\frac{1}{2}} t^{3 / 2}, {(\frac{P t}{m})}^{\frac{1}{2}}$

C.

${(\frac{8 P}{9 m})}^{\frac{1}{2}} t^{3 / 2}, {(\frac{2 P t}{m})}^{\frac{1}{2}}$

D.

${(\frac{8 P}{9 m})}^{\frac{1}{2}} t^{3 / 2}, {(\frac{P t}{m})}^{\frac{3}{2}}$

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Explanation

Velocity

As Fv= P= constant

i.e. $m \frac{d v}{d t} v = P$
$[A s F = \frac{m d v}{d t}]$
or $\int v d v = \int \frac{P}{m} d t$

By integrating both sides we get $\frac{v^{2}}{2} = \frac{P}{m} t + C_{1}$

As initially the body is at rest i.e. v= 0 at t= 0, so $C_{1} = 0 ∴ v = {(\frac{2 P t}{m})}^{1 / 2}$

Position

From the above expression $v = {(\frac{2 P t}{m})}^{1 / 2}$

or $\frac{d s}{d t} = {(\frac{2 P t}{m})}^{1 / 2}$ $[A s v = \frac{d s}{d t}]$

i.e. $\int d s = \int {(\frac{2 P t}{m})}^{1 / 2} d t$

By integrating both sides we get $s = {(\frac{2 P}{m})}^{1 / 2} . \frac{2}{3} t^{3 / 2} + C_{2}$

Now as at t= 0, s= 0, so $C_{2} = 0 ∴ s = {(\frac{8 P}{9 m})}^{1 / 2} t^{3 / 2}$

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A constant force F is applied on a body. The power (P) generated is related to the time elapsed (t) as

[This question is only for Dropper and XII batch]

A.

$P \propto t^{2}$

B.

$P \propto t$

C.

$P \propto \sqrt{t}$

D.

$P \propto t^{3 / 2}$

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Explanation

$F = \frac{m d v}{d t} ∴ F d t = m d v \Rightarrow v = \frac{F}{m} t$

Now $P = F \times v = F \times \frac{F}{m} t = \frac{F^{2} t}{m}$

If force and mass are constants then $P \propto t$

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A rod of length L is placed along the x-axis between x= 0 and x= L. The linear density (mass/length) $λ$ of the rod varies with the distance x from the origin as $λ$ = Rx. Here, R is a positive constant. Find the position of centre of mass of this rod.

[This question is only for Dropper and XII batch]

A.

$[0, \frac{2 L}{3}, 0]$

B.

$[0, 0, \frac{2 L}{3}]$

C.

$[\frac{2 L}{3}, 0, 0]$

D.

$[\frac{L}{3}, 0, 0]$

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The gravitational field due to a mass distribution is given by $\vec{I} = \frac{k}{x^{2}} \hat{i}$ , where k is a constant. Assuming the potential to be zero at infinity, find the potential at a point x = a. [This question is only for Dropper and XII batch]

A.

$\frac{k}{a^{2}}$

B.

$\frac{- k}{a^{2}}$

C.

$\frac{k}{a}$

D.

$\frac{- k}{a}$

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Explanation

We know,

$d V = - \vec{I} . \vec{d r} \Rightarrow d V = - \frac{k}{x^{2}} d x \Rightarrow \int d V = - k \int \frac{d x}{x^{2}} \Rightarrow V = \frac{k}{x} + C$

When $x = \infty, V = 0 \Rightarrow C = 0$

$\Rightarrow V = \frac{k}{x} ∴ A t x = a, V = \frac{k}{a}$

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The upper edge of a gate in a dam runs along water surface. The gate is 2 m high and 3 m wide and is hinged along a horizontal line through its center. Calculate the torque about hinge. [This question is only for Dropper and XII batch]

A.

$10^{4} N m$

B.

$2 \times 10^{4} N m$

C.

$3 \times 10^{4} N m$

D.

$4 \times 10^{4} N m$

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Explanation

The torque acting on the gate is given by the product of the force acting on it and the perpendicular distance between the line of action of the force and the hinge. The force acting on the gate is the weight of the gate, which is equal to its mass times the acceleration due to gravity.

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The specific heat of a substance varies with temperature t( $° C$ ) as
c= 0.20 + 0.14 t + 0.023 $t^{2} (c a l / g m ° C)$ . Find the heat required to raise the temperature of 2 gm of substance from 5 $° C$ to 15 $° C$ . [This question is only for Dropper and XII batch]

A.

41 cal

B.

82 cal

C.

80 cal

D.

40 cal

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Explanation

Heat required to raise the temperature of m gm of substance by dT is given as

dQ= mc dT $\Rightarrow Q = \int m c d T$

$∴$ To raise the temperature of 2 gm of substance from 5 $° C$ to 15 $° C$ is

$Q = {}_{5}^{15}\int 2 \times (0.2 + 0.14 t + 0.023 t^{2}) d T = 2 \times {[0.2 t + \frac{0.14 t^{2}}{2} + \frac{0.023 t^{3}}{3}]}_{5}^{15} = 82 c a l o r i e$

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An isolated container at 127 $° C$ contains an ice cube of mass 100 g at 0 $° C$ . The specific heat C of container varies with temperature according to relation C= a+bT, where a= 0.1 kcal/kg-K and b= 40 m cal/kg K. Find the mass of container, if the final temperature of container is 300 K.
[Take $L_{F}$ = 80 cal/g and specific heat of water 1 cal/g K]

[This question is only for Dropper and XII batch]

A.

500 g

B.

720 g

C.

940 g

D.

1200 g

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Explanation

Given specific heat of container, C= a+bT

The initial temperature of container $T_{i}$ = 127 + 273 = 400 K

and final temperature of container $T_{f}$ = 300 K

$.^{.} . H e a t l o s t b y c o n t a i n e r = - {}_{T_{i}}^{T_{f}}\int m_{C} (a + b T) d T (w h e r e m_{C} \to m a s s o f c o n t a i n e r)$

Heat lost= $- {}_{400}^{300}\int m_{C} (a + b T) d T$

= $- m_{C} {[a T + \frac{b T^{2}}{2}]}_{400}^{300}$

$= - m_{C} [(a 300 - a 400) + \frac{b}{2} ((300)^{2} - (400)^{2})] = m_{C} [100 a + 35000 b] = m_{C} [100 \times 0.1 \times 10^{3} + 35000 \times 40 \times 10^{- 3}] = m_{C} [10000 + 1400] = m_{C} 11400$

Heat gained by ice = $m_{i c e} L_{F} + m_{i c e} C_{w a t e r} ∆ T$

= $(0.1 \times 80 \times 10^{3}) + (0.1 \times 10^{3} \times 27) = 8000 + 2700$

= 10700 cal

From principle of calorimetry

Heat lost by container= Heat gained by ice

$.^{.} .$ $11400 m_{C} = 10700 m_{C} = \frac{107}{114} = 0.939 k g = 939 g$

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The temperature of n moles of an ideal gas is increased from $T_{0} t o 2 T_{0}$ through a process $P = \frac{α}{T}$ . Find the work done by the gas. [This question is only for Dropper and XII batch]

A.

$\frac{n R T_{0}}{2}$

B.

$n R T_{0}$

C.

$\frac{3}{2} n R T_{0}$

D.

2 $n R T_{0}$

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Explanation

PV= nRT (ideal gas equation) .....(i)

and $P = \frac{α}{T}$ .....(ii)

Divinding (i) by (ii), we get $V = \frac{n R T^{2}}{α} o r d V = \frac{2 n R T}{α} d T$

$.^{.} . W = \int_{V_{i}}^{V_{f}} P d V = \int_{T_{0}}^{2 T_{0}} (\frac{α}{T}) (\frac{2 n R T}{α}) d T = 2 n R T_{0}$

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