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UPSC NDA I 2024 General Ability Paper

Question : 83 of 150

Marks: +1, -0

N_{1}

W_{1}

N_{2}

W_{2}

g = 10.0\,\mathrm{m}\,\mathrm{s}^{-2}

Solution:

Concept:

The meter scale is in static equilibrium under its weight and the normal forces from the wedges.

Explanation:

Let the scale length be

1.0\,\mathrm{m}

and its mass

m = 0.24\,\mathrm{kg}

Weight

mg = 0.24 \times 10 = 2.4\,\mathrm{N}

acts at the midpoint, i.e.

0.5\,\mathrm{m}

from either end.

Wedge

W_1

is at

0.2\,\mathrm{m}

from the left end.

Wedge

W_2

is at

0.4\,\mathrm{m}

from the right end, so its distance from the left end is

1.0 - 0.4 = 0.6\,\mathrm{m}

Distance between

W_1

and

W_2

0.6 - 0.2 = 0.4\,\mathrm{m}

Apply vertical equilibrium:

N_1 + N_2 = mg = 2.4\,\mathrm{N}

(Equation 1).

Take torques about the centre of mass (at

0.5\,\mathrm{m}

from left).

Distance of

W_1

from centre:

0.5 - 0.2 = 0.3\,\mathrm{m}

(clockwise torque).

Distance of

W_2

from centre:

0.6 - 0.5 = 0.1\,\mathrm{m}

(anticlockwise torque).

For rotational equilibrium:

N_1 \times 0.3 = N_2 \times 0.1

Thus

3N_1 = N_2

(Equation 2).

Substitute Equation 2 into Equation 1:

N_1 + 3N_1 = 2.4

4N_1 = 2.4 \Rightarrow N_1 = 0.6\,\mathrm{N}

Then

N_2 = 2.4 - 0.6 = 1.8\,\mathrm{N}

Answer:

N_1 = 0.6\,\mathrm{N}

and

N_2 = 1.8\,\mathrm{N}

. This corresponds to option C.

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