KCET · Physics · Laws of Motion
A mass of \(10 \mathrm{~kg}\) is suspended from a spring balance. It is pulled aside by a horizontal string so that it makes an angle of \(60^{\circ}\) with the vertical. The new reading of the balance is
- A \(20 \mathrm{~kg}-\mathrm{wt}\)
- B \(10 \mathrm{~kg}\)-wt
- C \(10 \sqrt{3} \mathrm{~kg}-\mathrm{wt}\)
- D \(20 \sqrt{3} \mathrm{~kg}\)-wt
Answer & Solution
Correct Answer
(A) \(20 \mathrm{~kg}-\mathrm{wt}\)
Step-by-step Solution
Detailed explanation
The situation is shown in figure
At an angle of \(60^{\circ}\),
\(\mathrm{T} \cos \theta =\mathrm{mg} \)
\(\mathrm{T} =\frac{\mathrm{mg}}{\cos \theta}=\frac{10 g}{\cos 60^{\circ}} \)
\( =\frac{10}{1 / 2} \mathrm{~kg}-\mathrm{wt} . \)
\(=20 \mathrm{~kg}-\mathrm{wt}\)
At an angle of \(60^{\circ}\),
\(\mathrm{T} \cos \theta =\mathrm{mg} \)
\(\mathrm{T} =\frac{\mathrm{mg}}{\cos \theta}=\frac{10 g}{\cos 60^{\circ}} \)
\( =\frac{10}{1 / 2} \mathrm{~kg}-\mathrm{wt} . \)
\(=20 \mathrm{~kg}-\mathrm{wt}\)
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