JEE Main · 2020 · 05 Sep 2020 · Shift 1 · Rotational Mechanics

A force

F = (\hat{i} + 2 \hat{j} + 3 \hat{k})

N acts at a point

(4 \hat{i} + 3 \hat{j} - \hat{k})

m. Then the magnitude of torque about the point

(\hat{i} + 2 \hat{j} + \hat{k})

m will be

x

N-m. The value of

x

is ______.

Answer: 195

Solution

The torque about the chosen point is

τ = r \times F

, where

r

runs from that point to the point of application:

r = (4 \hat{i} + 3 \hat{j} - \hat{k}) - (\hat{i} + 2 \hat{j} + \hat{k}) = 3 \hat{i} + \hat{j} - 2 \hat{k}

. With

F = \hat{i} + 2 \hat{j} + 3 \hat{k}

, the cross product is

τ = (1 \cdot 3 - (- 2) \cdot 2) \hat{i} - (3 \cdot 3 - (- 2) \cdot 1) \hat{j} + (3 \cdot 2 - 1 \cdot 1) \hat{k} = 7 \hat{i} - 11 \hat{j} + 5 \hat{k}

. Its magnitude is

7^{2} + 1 1^{2} + 5^{2} = 195

\cdot

m, so

x = 195

JEE Main · 2020 · 05 Sep 2020 · Shift 1 · Integer answer

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