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### JEE Main 2016 (Online) 9th April Morning Slot

A car of weight W is on an inclined road that rises by 100 m over a distance of 1 km and applies a constant frictional force ${W \over 20}$ on the car. While moving uphill on the road at a speed of 10 ms−1, the car needs power P. If it needs power ${p \over 2}$ while moving downhill at speed v then value of $\upsilon$ is :
A
20 ms$-$1
B
15 ms$-$1
C
10 ms$-$1
D
5 ms$-$1

## Explanation

Here, tan$\theta$ = ${{100} \over {1000}} = {1 \over {10}}$

$\therefore$   sin$\theta$ = ${1 \over {10}}$ (as   $\theta$  is very small),

when car is moving uphill : P = f $\times$ u

=  (wsin$\theta$ + f) $\times$ u

=  $\left( {{w \over {10}} + {w \over {20}}} \right) \times 10$

P = ${{3w} \over {20}} \times 10$ = ${{3w} \over 2}$

When car is moving down hill : $\therefore$   ${P \over 2}$ = (wsin$\theta$ $-$ f) $\times$ v

$\Rightarrow$   ${{3w} \over 4}$ = $\left( {{w \over {10}} - {w \over {20}}} \right)$ $\times$ v

$\Rightarrow$   ${{w \over {20}} \times }$ v = ${{3w} \over 4}$

$\Rightarrow$   v = 15 m/s
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### 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
3

### 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
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### 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