Category Archives: Physics

Water in Vertical Circle

Consider a small bucket full of water tied to a string whirled around in vertical circle of radius r without water falling down. At the topmost position when the speed of the inverted bucket is v,

(A) $\frac{{{v^2}}}{r} = g$ necessarily
(B) $\frac{{{v^2}}}{r} < g$ necessarily
(C) $\frac{{{v^2}}}{r} > g$ necessarily
(D) None of the options given

Solution

Consider water as the system. Let, m be the mass of water. The weight mg acts downwards and the normal reaction N also acts downwards.

$mg + N = \frac{{m{v^2}}}{r}$

$ \Rightarrow \frac{{m{v^2}}}{r} \ge mg$

$ \Rightarrow \frac{{{v^2}}}{r} \ge g$

Hence, Option (D).

Weighing Machine in Elevator

A man inside an elevator uses weighing machine to weigh himself. With what acceleration should the elevator descend so that the weighing machine reports the weight of the man to be half of its true value?

The reading of the weighing machine depends on the force by which the machine is pressed. Let us say the machine is pressed downwards by force N. The machine will exert an equal upward force N on the man.

Let m be the mass of the man. For the man system,

mg – N = ma

$mg – \frac{{mg}}{2} = ma$

$ \Rightarrow a = \frac{g}{2}$

Pulley + Block + Rope

A uniform rope of linear mass density $\lambda $ is used to release block m with uniform acceleration a. Find the tension at a point P on the rope at a distance l from the block as shown in the figure.

Solution

Let mass m and the rope of length l above the block be the system.

Mass of rope of length $l = \lambda l$

For the system under consideration, $(m + \lambda l)g – {T_P} = (m + \lambda l)a$

$ \Rightarrow {T_P} = (m + \lambda l)(g – a)$

Conducting Bar Across $R_1$ & $R_2$

A conducting bar PQ is free to slide on two parallel conducting rails as shown in figure. Two resistors $R_1$ and $R_2$ are connected across the ends of the rails. There is a uniform magnetic field B pointing into the page. An external agent pulls the bar to the left at a constant speed v. The correct statement about the directions of induced currents $I_1$ and $I_2$ flowing through $R_1$ and $R_2$ respectively is:

(A) $I_1$ is in anticlockwise direction and $I_2$ is in clockwise direction
(B) Both $I_1$ and $I_2$ are in anticlockwise direction
(C) Both $I_1$ and $I_2$ are in clockwise direction
(D) $I_1$ is in clockwise direction and $I_2$ is in anticlockwise direction

Solution

As PQ moves leftwards, the area of loop ABQP decreases causing reduction in flux. So, the direction of current through $R_1$ should be such that the flux can be increased. An upward current through $R_1$ helps to strengthen the magnetic field. Upward current through $R_1$ is the clockwise current through the left loop.

As PQ moves leftwards, the area of loop A’B’QP increases causing increase in flux. So, the direction of current through $R_2$ should be such that the flux can be decreased. An upward current through $R_2$ helps to weaken the magnetic field. Upward current through $R_2$ is the anticlockwise current through the right loop.

Hence, Option (D).

Centripetal & Centrifugal Forces

In circular motion, why don’t centripetal and centrifugal forces cancel each other? Well, they don’t cancel each other because they are not present simultaneously for the same observer. Only one is present at a time for one observer. If you take ground observer – inertial observer to be more precise – then only centripetal force is applicable and of course it is applicable towards the centre. There is no centrifugal force present here. Since both are not present simultaneously, they don’t cancel each other. In case of non-inertial or accelerated observer, only centrifugal force is present – it acts away from the centre. There is no centripetal force applicable. Since both are not present simultaneously, they can’t cancel each other.