2 Marks Questions

Question 12 Marks

Is it true that the reaction of a gravitational force is always gravitational, of an electromagnetic force is always electromagnetic and so on?

Answer

Inertia pushes us forward.
As we are moving with the same momentum as of the car when driver applies breaks the car stops but our body keeps moving in same inertia as of car thus we are pushed forward.

View full question & answer→

Question 22 Marks

List all the forces acting on:

The pulley A.
The boy and.
The block C in figure.

Answer

Torque (angular force) is acting on pulley A.
Gravitational force is acting on boy as well pull due to weight of block C.
On block C gravity as well force upward due to force applied by boy are acting.

View full question & answer→

Question 32 Marks

A body builder exerts a force of 150N against a bullworker and compresses it by 20cm. Calculate the spring constant of the spring in the bullworker.

Answer

Force exerted by the body builder against the bullworker = 150N

Compression in the bullworker, x = 20cm = 0.2m

$\therefore$ Total force exerted, f = kx = 150

Here, k is the spring constant of the spring in the bullworker.

$\therefore\text{k}=\frac{150}{0.2}=\frac{1500}{2}=750\text{N/m}$

Hence, the spring constant of the spring in the bullworker is 750N/m.

View full question & answer→

Question 42 Marks

A monkey is sitting on a tree limb. The limb exerts a normal force of 48N and a frictional force of 20N. Find the magnitude of the total force exerted by the limb on the monkey.

Answer

Given: The limb of the tree exerts a normal force of 48N and a frictional force of 20N.

So, resultant magnitude of the force if given by

$\text{R}=\sqrt{(48^2+20^2)}$

$=\sqrt{2304+400}$

$=\sqrt{2704}=52\text{N}$

$\therefore$ The magnitude of the total force exerted by the limb on the monkey is 52N.

View full question & answer→

Question 52 Marks

The force with which the earth attracts an object is called the weight of the object. Calculate the weight of the moon from the following data: The universal constant of gravitation G = 6.67 × 10^-11N-m²/kg², mass of the moon = 7.36 × 10²²kg, mass of the earth = 6 × 10²⁴kg and the distance between the earth and the moon = 3.8 × 10⁵km.

Answer

The force between the Earth and the Moon is given by $\text{F}=\frac{\text{GMm}}{\text{r}^2}.$

Here, M is the mass of the earth; m is the mass of the moon and r is the distance between Earth and Moon.

On substituting the values, we get:

$\text{F}=\frac{6.67\times10^{-11}\times7.36\times10^{22}\times6\times10^{24}}{3.8\times3.8\times10^{16}}$

$=\frac{6.67\times7.36\times10^{35}}{(3.8)^2\times10^{16}}$

$=20.3\times10^{19}=2.03\times10^{20}$

$\approx2.0\times10^{20}\text{N}$

$\therefore$ The weight of the moon is $2.0\times10^{20}\text{N}.$

View full question & answer→

Question 62 Marks

Find the ratio of the magnitude of the electric force to the gravitational force acting between two protons.

Answer

Charge of the proton, q = 1.6 × 10^-19C

Mass of the proton = 1.67 × 10^-27kg

Let the distance between two protons be r.

Coulomb force (electric force) between the protons is given by

$\text{f}_{\text{e}}=\frac{1}{4\pi\in_{0}}\times\frac{\text{q}^2}{\text{r}^2}$

$=\frac{9\times10^9\times(1.6)^2\times10^{-38}}{\text{r}^2} $

Gravitational force between the protons is given by

$\text{f}_{\text{g}}=\frac{\text{Gm}^2}{\text{r}^2}$

$=\frac{6.67\times10^{-11}\times(1.67\times10^{-27})^2}{\text{r}^2}$

On dividing f_e by fgfe by f_g, we get:

$\frac{\text{f}_{\text{e}}}{\text{f}_{\text{g}}}=\frac{1}{4\pi\in_0}\times\frac{\text{q}^2}{\text{r}^2}\times\frac{\text{r}^2}{\text{Gm}^2}$

$=\frac{9\times10^9\times1.6\times1.6\times10^{-38}}{6.67\times10^{-11}\times1.67\times1.67\times10^{-54}}$

$=\frac{9\times(1.6)^2\times10^{-29}}{6.67\times(1.67)^2\times10^{-65}}$

$=1.24\times10^{36}$

View full question & answer→

Question 72 Marks

A body of mass m is placed on a table. The earth is pulling the body with a force mg. Taking this force to be the action what is the reaction?

Answer

No, The increase acceleration will apply on both sides of beam so there would be no profit such as to weigh moongfali more than actual.
(in case it was electronic wieghing machine the moongfali would weigh more definitely)

View full question & answer→

Question 82 Marks

Figure shows a boy pulling a wagon on a road. List as many forces as you can which are relevant with this figure. Find the pairs of forces connected by Newton's third law of motion.

Answer

Pull Weight of wagon And friction may be acting on the wagon.
Friction + weight of wagon are cancelled out by pulling force of boy.

View full question & answer→

Question 92 Marks

List all the forces acting on the block B in figure.

Answer

As inertial frame is used to understand unknown forces acting on body under motion, as in case of car when breaks are applied no apparent force is applied but still our body moves forward so to understand this frame of reference is used but actually this does not exit.

View full question & answer→

Question 102 Marks

A boy is sitting on a chair placed on the floor of a room. Write as many action-reaction pairs of forces as you can.

Answer

In free fall there would be no force exerted on boy in case mass of box is equal to or less than mass of boy. but in case mass of box is greater than mass of boy in that cases there would be a little force exerted on boys head. But when boy will strike ground the force attained by box that is Mg would be exerted on boy's head and that would be way too much greater than the weight while free falling.

View full question & answer→

Physics 2 Marks Questions

Test yourself on this topic