KCET · Physics · Rotational Motion
A door \(1.6 \mathrm{~m}\) wide requires a force of \(1 \mathrm{~N}\) to be applied at the free end to open or close it. The force that is required at a point \(0.4 \mathrm{~m}\) distance from the hinges for opening or closing the door is
- A \(1.2 \mathrm{~N}\)
- B \(3.6 \mathrm{~N}\)
- C \(2.4 \mathrm{~N}\)
- D \(4 \mathrm{~N}\)
Answer & Solution
Correct Answer
(D) \(4 \mathrm{~N}\)
Step-by-step Solution
Detailed explanation
Here, torque \(\tau=1.6 \times 1=1.6 \mathrm{~N}-\mathrm{m}\) So, when \(d=0.4 \mathrm{~m}\),
\(\mathrm{F}=\frac{\tau}{\mathrm{d}}=\frac{1.6}{0.4}=4 \mathrm{~N}\)
\(\mathrm{F}=\frac{\tau}{\mathrm{d}}=\frac{1.6}{0.4}=4 \mathrm{~N}\)
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