1
MCQ (Single Correct Answer)

JEE Main 2016 (Online) 10th April Morning Slot

A particle of mass M is moving in a circle of fixed radius R in such a way that its centripetal acceleration at time t is given by n2 R t2 where n is a constant. The power delivered to the particle by the force acting on it, is :
A
M n2 R2 t
B
M n R2 t
C
M n R2 t2
D
$${1 \over 2}$$ M n2 R2 t2

Explanation

We know,

centripetal acceleration = $${{{V^2}} \over R}$$

$$ \therefore $$   According to question,

$${{{V^2}} \over R}$$ = $${n^2}R{t^2}$$

$$ \Rightarrow $$   V2 = n2 R2 t2

$$ \Rightarrow $$   V = nRt

$$ \Rightarrow $$   $${{dV} \over {dt}}$$ = nR

Power (P) = Force (F) $$ \times $$ Velocity (V)

= M $${{dV} \over {dt}}$$(V)

= M (nR) (nRt)

= Mn2R2t
2
MCQ (Single Correct Answer)

JEE Main 2016 (Online) 10th April Morning Slot

A neutron moving with a speed ‘v’ makes a head on collision with a stationary hydrogen atom in ground state. The minimum kinetic energy of the neutron for which inelastic collision will take place is :
A
10.2 eV
B
16.8 eV
C
12.1 eV
D
20.4 eV

Explanation

Let, velocity offer collision = v1

$$ \therefore $$   From conservation of momentum,

mv = (m + m) v1

$$ \Rightarrow $$    v1 = $${v \over 2}$$

$$ \therefore $$   Loss in kinetic energy

= $${1 \over 2}$$ mv2 $$-$$ $${1 \over 2}$$ (2m) $$ \times $$ $${\left( {{v \over 2}} \right)^2}$$

= $${1 \over 4}\,$$mv2

lost kinetic energy is used by the electron to jump from first orbit to second orbit.

$$ \therefore $$   $${1 \over 4}$$mv2 = (13.6 $$-$$ 3.4) eV = 10.2 eV

$$ \Rightarrow $$  $${1 \over 2}$$mv2 = 20.4 eV
3
MCQ (Single Correct Answer)

JEE Main 2016 (Online) 10th April Morning Slot

Velocity-time graph for a body of mass 10 kg is shown in figure. Work-done on the body in first two seconds of the motion is :

A
12000 J
B
$$-$$ 12000 J
C
$$-$$ 4500 J
D
$$-$$ 9300 J

Explanation



Here u = 50 m/s , what t = 0

$$\alpha $$  =  $${{\Delta v} \over {\Delta t}}$$  =  $${{50 - 0} \over {0 - 10}}$$  =  $$-$$5 m/s2

Speed of the body at t = 2 s

v   =   u + at

=  50 + ($$-$$ 5) $$ \times $$ 2

=  40 m/s

From work energy theorem,

$$\Delta $$w  =  $${1 \over 2}m{v^2} - {1 \over 2}m{u^2}$$

=  $${1 \over 2}$$ m(v2 $$-$$u2)

= $${1 \over 2}$$ $$ \times $$ 10 $$ \times $$ (402 $$-$$ 502)

=  5 $$ \times $$ ($$-$$10)(90)

=  $$-$$ 4500 J
4
MCQ (Single Correct Answer)

JEE Main 2017 (Offline)

A body of mass m = 10–2 kg is moving in a medium and experiences a frictional force F = –kv2. Its initial speed is v0 = 10 ms–1. If, after 10 s, its energy is $${1 \over 8}mv_0^2$$, the value of k will be:
A
10-1 kg m-1 s-1
B
10-3 kg m-1
C
10-3 kg s-1
D
10-4 kg m-1

Explanation

According to the question, final kinetic energy = $${1 \over 8}mv_0^2$$

Let final speed of the body = Vf

So final kinetic energy = $${1 \over 2}mv_f^2$$

According to question,

$${1 \over 2}mv_f^2$$ = $${1 \over 8}mv_0^2$$

$$ \Rightarrow {v_f} = {{{v_0}} \over 2}$$ = $${{10} \over 2}$$ = 5 m/s

Given that, F = –kv2

$$ \Rightarrow $$ $$m\left( {{{dv} \over {dt}}} \right)$$$$ = - k{v^2}$$

$$ \Rightarrow {10^{ - 2}}\left( {{{dv} \over {dt}}} \right) = - k{v^2}$$

$$ \Rightarrow \int\limits_{10}^5 {{{dv} \over {{v^2}}}} = - 100k\int\limits_0^{10} {dt} $$

$$ \Rightarrow {1 \over 5} - {1 \over {10}} = 100k \times 10$$

$$ \Rightarrow k = {10^{ - 4}}kg\,{m^{ - 1}}$$

Questions Asked from Work Power & Energy

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