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Physics NEET Questions — Free MCQ Practice

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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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Electric field is given by E $= \frac{100}{x^{2}}$ . Find the potential difference between x= 10 and x= 20 m. [This question is only for Dropper and XII batch]

A.

5 V

B.

10 V

C.

15 V

D.

20 V

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Explanation

$E = - \frac{d V}{d x} \Rightarrow d V = - E d x \Rightarrow {}_{A}^{B}\int d V = - {}_{A}^{B}\int E . d x \Rightarrow V_{B} - V_{A} = - {}_{10}^{20}\int \frac{100}{x^{2}} = - 5 v o l t s P o t e n t i a l d i f f e r e n c e = 5 v o l t .$

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The electric field vector in a region given by $\vec{E} = (3 \hat{i} + 4 y \hat{j}) V m^{- 1}$ . Calculate the potential at (1m, 1m) taking potential at origin to be zero. [This question is only for Dropper and XII batch]

A.

5V

B.

3V

C.

-1V

D.

-5V

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Explanation

Here, $\vec{E} = \hat{i} + 4 y \hat{j}, \vec{r} = 0 \hat{i} + 0 \hat{j}, \vec{r_{2}} = \hat{i} + \hat{j}, d r = d x \hat{i} + d y \hat{j}, V_{1} = 0, V_{2} = V$

$\Rightarrow V - 0 = - \int_{T 1}^{T 2} (3 \hat{i} + 4 y \hat{j}) . (d x \hat{i} + d y \hat{j}) \Rightarrow V - 0 = - {}_{0}^{1}\int 3 d x - {}_{0}^{1}\int 4 y d y \Rightarrow V = - 3 - 4 {[\frac{y^{2}}{2}]}_{0}^{1} = - 3 - \frac{4}{2} = - 5 V$

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Each of a parallel-plate air capacitor has an area S. What amount of work has to be performed to slowly increase the distance between the plates from $x_{1} t o x_{2}$ . If the voltage across the capacitor, which is equal to V, is kept constant in the process. [This question is only for Dropper and XII batch]

A.

$\frac{ε_{0} S V}{2}$

B.

$ε_{0} S V^{2} [\frac{1}{x_{1}} - \frac{1}{x_{2}}]$

C.

$\frac{ε_{0} S V^{2}}{2} [\frac{1}{x_{1}} - \frac{1}{x_{2}}]$

D.

$ε_{0} S V$

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Explanation

When voltage is kept constant, the force acting on each plate of capacitor will depend on the distance between the plates.

So, elementary work done by agent, in its displacement over a distance dx, relative to the other,
dW= $- F_{x} d x$

But, $F_{x} = - (\frac{σ (x)}{2 ε_{0}}) S σ (x) a n d σ (x) = \frac{ε_{0} V}{x}$

Hence, $W = \int d W = {}_{x 1}^{x 2}\int \frac{1}{2} ε_{0} \frac{S V^{2}}{x^{2}} d x = \frac{ε_{0} S V^{2}}{2} [\frac{1}{x_{1}} - \frac{1}{x_{2}}]$

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Using the concept of energy density, find the total energy stored by a shell of radius R and charge Q. [This question is only for Dropper and XII batch]

A.

$\frac{Q^{2}}{{πε}_{0} R}$

B.

$\frac{Q^{2}}{2 {πε}_{0} R}$

C.

$\frac{Q^{2}}{4 {πε}_{0} R}$

D.

$\frac{Q^{2}}{8 {πε}_{0} R}$

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Six particles situated at the corners of a regular hexagon of side a move at constant speed v. Each particle maintains a direction towards the particle at the next. The time which the particle will take to meet each other is

A.

$\frac{2 a}{v} \sec$

B.

$\frac{a}{v} \sec$

C.

$\frac{2 a}{3 v} \sec$

D.

$\frac{3 a}{v} \sec$

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A particle is thrown upwards from ground. It experiences a constant resistance force which can produce retardation of 2 $m / s^{2}$ . The ratio of time of ascent to the time of descent is: $[g = 10 m / s^{2}]$

A.

1:1

B.

$\sqrt{\frac{2}{3}}$

C.

$\frac{2}{3}$

D.

$\sqrt{\frac{3}{2}}$

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A bullet loses $\frac{1}{20}$ of its velocity passing through a plank. The least number of planks required to stop the bullet is (All planks offers same retardation)

A.

10

B.

11

C.

12

D.

23

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Explanation

For a bullet to be stopped, it must lose all its velocity. If it loses 1/20th of its velocity after passing through one plank, then to lose its entire velocity, it needs to pass through 20 planks. Therefore, the least number of planks required to stop the bullet is 11.

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A body starts from the origin and moves along the X-axis such that the velocity at any instant is given by $(4 t^{3} - 2 t)$ , where t is in sec and velocity in m/s. What is the acceleration of the particle, when it is 2 m from the origin ?

A.

28 m/s²

B.

22 m/s²

C.

12 m/s²

D.

10 m/s²

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Explanation

$v = 4 t^{3} - 2 t$ (given) ∴ $a = \frac{d v}{d t} = 12 t^{2} - 2$

and $x = \int_{0}^{t} v d t = \int_{0}^{t} (4 t^{3} - 2 t) d t = t^{4} - t^{2}$

When particle is at 2m from the origin $t^{4} - t^{2} = 2$

⇒ $t^{4} - t^{2} - 2 = 0$ ⇒ $(t^{2} - 2) (t^{2} + 1) = 0$ ⇒ $t = \sqrt{2} \sec$

Acceleration at $t = \sqrt{2} \sec$ given by,

$a = 12 t^{2} - 2 = 12 \times 2 - 2$ = $22 m / s^{2}$

Physics MCQs for NEET — Practice Questions with Answers

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