For the arrangement shown in the figure, the tension in the string is given by $(s i n 37^{\circ} = \frac{3}{5})$ . Take $g = 10 m s^{- 2}$

30 N

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40 N

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60 N

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30 N

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Solution

The correct option is B $40 N$

Net pulling force on the system $F = 10 g s i n 37^{\circ} - 4 g = 20 N$
maximum force of friction
$f_{m a x} = μ mg cos 37^{\circ}$
$= 0.7 \times 10 \times 10 \times \frac{4}{5} = 56 N$
Since $F < f_{m a x}$ , system will not move. Equilibrium of $4 k g$ gives $T = 40 N$ .

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For the arrangement shown in the figure, the tension in the string is given by (sin37∘=35). Take g=10 ms−2

The correct option is B 40 N Net pulling force on the system F=10gsin37∘−4g=20 N maximum force of friction fmax=μmg cos37∘ =0.7×10×10×45=56 N Since F<fmax, system will not move. Equilibrium of 4 kg gives T=40 N.

For the arrangement shown in the figure, the tension in the string is given by $(s i n 37^{\circ} = \frac{3}{5})$ . Take $g = 10 m s^{- 2}$