Test Series

Take Zigya Full and Sectional Test Series. Time it out for real assessment and get your results instantly.

Test Yourself

Practice and master your preparation for a specific topic or chapter. Check you scores at the end of the test.

JEE Physics Solved Question Paper 2016

Multiple Choice Questions

The one which does not represent a force in any context is

Friction
Impulse
Tension
Weight

The work-energy theorem states that the change in

kinetic energy of a particle is equal to the work done on it by the net force
kinetic energy of a particle is equal to the work done by one of the forces acting on it.
potential energy of a particle is equal to the work done on it by the net force
potential energy of a particle is equal to the work done by one of the forces acting on it

A car of masses 1500 kg is lifted up a distance of 30 m by crane A in 0.5 minutes. The second crane B does the same job in 1 minute. The ratio of their powers is

1 : 2
2 : 1
1 : 1
1 : 4

Water from a hose pipe of radius 5 cm strikes a wall normally at a speed of 5 ms^-1. The force exerted on the wall in newton is

$13.5 π$
$6.25 π$
$62.5 π$
$27 π$

The position vectors of two identical particles with respect to the origin in three-dimensional coordinate system are r₁ and r₂. The position vector of centre of mass of the system is given by

r₁ + r₂
2(r₁ + r₂)
r₁ − r₂
$\frac{r_{1} + r_{2}}{2}$

If a body of moment of inertia 2 kg m² revolves about its own axis making 2 rotations per second, then its angular momentum (in Js) is

$2 π$
$4 π$
$6 π$
$8 π$

A rigid body is the one in which

it can have only rotational motion
it can have only translational motion
the distances between all pairs of particles of the body do not change
its centre of mass always lies inside the material of the body

8.
A body hanging from a massless spring stretches it by 3 cm on Earth's surface. At a place 800 km above the Earth's surface, the same body will stretch the spring by (Radius of Earth = 6400 km)

$(\frac{34}{27}) cm$

$(\frac{64}{27}) cm$

$(\frac{27}{64}) cm$

$(\frac{27}{34}) cm$

$(\frac{64}{27}) cm$

Acceleration due to gravity,

$g = \frac{GM}{r^{2}} ∴ g \propto \frac{1}{r^{2}}$

When the body is hanged on a spring

F = − kx = mg ..... (i)

acts on it, where x is extension in spring. Let 800 km above the Earth's surface, the stretch in the length of spring is x' and value of g is g'.

$g = \frac{GM}{R^{2}} and g' = \frac{GM}{(R + h)^{2}} So, g' = \frac{g}{{(1 + \frac{h}{R})}^{2}} . . . . . . (ii)$

where, R = radius of Earth

From Eqs.(i), we get

kx = mg ⇒ x ∝ g ...... (iii)

Therefore, $\frac{x'}{x} = \frac{g'}{g}$

From Eqs. (ii) and (iii), we get

$\frac{x'}{x} = \frac{g}{{(1 + \frac{h}{R})}^{2}} / g = \frac{g}{g {(1 + \frac{800 \times 10^{3}}{6400 \times 10^{3}})}^{2}} = \frac{g}{g {(1 + \frac{1}{8})}^{2}} = \frac{g}{g {(\frac{9}{8})}^{2}} = \frac{g \times 64}{g \times 81} = \frac{64}{81} x' = \frac{64}{81} x [x = 3 cm] \Rightarrow x' = \frac{64}{81} \times 3 = \frac{64}{27} \Rightarrow x' = \frac{64}{27} cm$