Laws of Motion

JEE Physics · 73 questions · Page 8 of 8 · Click an option or "Show Solution" to reveal answer

Q71

A block of mass

m

is placed on a surface with a vertical cross section given by

y = {{{x^3}} \over 6}.

If the coefficient of friction is

0.5,

the maximum height above the ground at which the block can be placed without slipping is:

{1 \over 6}m

{2 \over 3}m

{1 \over 3}m

{1 \over 2}m

Correct Answer

Option A

Solution

At limiting equilibrium,

\mu = \tan \theta

Equation of the surface,

y = {{{x^3}} \over 6}

Slope,

\tan \theta = \mu = {{dy} \over {dx}} = {{{x^2}} \over 2}

Given that, Coefficient of friction

\mu = 0.5

$\therefore$

\,\,\,\,0.5 = {{{x^2}} \over 2}

\Rightarrow \,\,\,x = \pm \,1

Now,

y = {{{x^3}} \over 6} = {1 \over 6}m

Q72

A small ball of mass m is thrown upward with velocity u from the ground. The ball experiences a resistive force mkv2 where v is its speed. The maximum height attained by the ball is :

{1 \over k}{\tan ^{ - 1}}{{k{u^2}} \over {2g}}

{1 \over {2k}}{\tan ^{ - 1}}{{k{u^2}} \over g}

{1 \over {2k}}\ln \left( {1 + {{k{u^2}} \over g}} \right)

{1 \over k}\ln \left( {1 + {{k{u^2}} \over {2g}}} \right)

Correct Answer

Option C

Solution

Fnet = ma $\Rightarrow$ -mg - mkv2 =

mv{{dv} \over {ds}}

$\Rightarrow$

ds = {{ - vdv} \over {g + k{v^2}}}

$\Rightarrow$

\int\limits_{s = 0}^{{H_{\max }}} {ds} = \int\limits_{v = u}^{v = 0} {{{ - vdv} \over {g + k{v^2}}}}

$\Rightarrow$ Hmax =

{1 \over {2k}}\ln \left( {{{g + k{u^2}} \over g}} \right)

{1 \over {2k}}\ln \left( {1 + {{k{u^2}} \over g}} \right)

Q73

At any instant the velocity of a particle of mass

500 \mathrm{~g}

\left(2 t \hat{i}+3 t^{2} \hat{j}\right) \mathrm{ms}^{-1}

. If the force acting on the particle at

t=1 \mathrm{~s}

(\hat{i}+x \hat{j}) \mathrm{N}

. Then the value of

x

will be:

A 2

B 4

C 6

D 3

Correct Answer

Option D

Solution

Given the velocity vector of a particle $v = (2t \hat{i}+3 t^{2} \hat{j}) \, \text{ms}^{-1}$ , the acceleration $a$ is the derivative of the velocity vector with respect to time.

So, we have: $a = \dfrac{dv}{dt} = (2 \hat{i} + 6t \hat{j}) \, \text{ms}^{-2}$ .

At $t=1 \, \text{s}$ , the acceleration $a$ is $(2 \hat{i} + 6 \hat{j}) \, \text{ms}^{-2}$ .

According to Newton's second law, the force $F$ is equal to the mass $m$ times acceleration $a$ .

The mass $m$ is given as $500 \, \text{g}$ , or equivalently, $0.5 \, \text{kg}$ .

Therefore, the force $F$ on the particle at $t=1 \, \text{s}$ is: $F = m \cdot a = 0.5 \cdot (2 \hat{i} + 6 \hat{j}) = (1 \hat{i} + 3 \hat{j}) \, \text{N}$ .

So, the force acting on the particle at $t=1 \, \text{s}$ is $(\hat{i} + x \hat{j}) \, \text{N}$ , where $x=3$ .

Therefore, the answer is $x=3$ .

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