Acceleration due to gravity is g on the surface of the earth. Then the value of the acceleration due to gravity at a height of 32 km above earth's surface is : (Assume radius of earth to be 6400 km)

• 0.99 g

• 0.8 g

• 1.01 g

• 0.9 g

According to Kepler's law of planetary motion if T represents time period and r is orbital radius, then for two planets these are related as

• ${\left(\frac{{\mathrm{T}}_1}{{\mathrm{T}}_2}\right)}^3 = {\left(\frac{{\mathrm{r}}_1}{{\mathrm{r}}_2}\right)}^2$

• ${\left(\frac{{\mathrm{T}}_1}{{\mathrm{T}}_2}\right)}^{3/2} = \frac{{\mathrm{r}}_1}{{\mathrm{r}}_2}$

• $\left(\frac{{\mathrm{T}}_1}{{\mathrm{T}}_2}\right) = {\left(\frac{{\mathrm{r}}_1}{{\mathrm{r}}_2}\right)}^3$

• $\left(\frac{{\mathrm{T}}_1}{{\mathrm{T}}_2}\right) = {\left(\frac{{\mathrm{r}}_1}{{\mathrm{r}}_2}\right)}^{3/2}$

The dimensions of kinetic energy is

• [M²L²T¹]

• [M¹L²T¹]

• [M¹L²T^-2]

• [M¹L²T^-1]

In an experiment to measure the height of a bridge by dropping stone into water underneath if the error in measurement of time is 0.1 s at the end of 2 s, then the error in estimation of height of bridge will be

• 0.49 m

• 0.98 m

• 1.37 m

• 1.96 m

If the figure below represents a parabola, identify the physical quantities representing Y and X for constant acceleration

• X = time, Y = velocity

• X = velocity, Y = time

• X = time, Y = displacement

• X = time, Y = acceleration

A ball which is at rest is dropped from a height h metre. As it bounces off the floor its speed is 80% of what it was just before touching the ground. The ball will then rise to nearly a height

• 0.94 h

• 0.80 h

• 0.75 h

• 0.64 h

Two bodies of different masses are dropped from heights of 16 m and 25 m respectively. The ratio of the time taken by them to reach the ground is

• $\frac{25}{16}$

• $\frac{5}{4}$

• $\frac{4}{5}$

• $\frac{16}{25}$

The recoil velocity of a 4.0 kg rifle that shoots a 0.050 kg bullet at a speed of 280 ms^-1 is

• + 3.5 ms^-1

• − 3.5 ms^-1

• $- \sqrt{\left(3.5\right)} {\mathrm{ms}}^{-1}$

• $+ \sqrt{\left(3.5\right)} {\mathrm{ms}}^{-1}$

A man of height h walks in a straight path towards a lamp post of height H with uniform velocity u. Then the velocity of the edge of the shadow on the ground will be

• $\frac{\mathrm{hu}}{\left(\mathrm{H} - \mathrm{h}\right)}$

• $\frac{\mathrm{Hu}}{\left(\mathrm{H} + \mathrm{h}\right)}$

• $\frac{\left(\mathrm{H} - \mathrm{h}\right)}{\mathrm{Hu}}$

• $\frac{\left(\mathrm{H} + \mathrm{h}\right)}{\mathrm{Hu}}$

A train of 150 m length is going towards north direction at a speed of 10 ms^-1. A parrot flies at a speed of 5 ms^-1 towards south direction parallel to the railway track. The time taken by the parrot to cross the train is equal to

• 12 s

• 8 s

• 15 s

• 10 s