A light string passes over a frictionless pulley. To one of its ends, a mass of 6 kg is attached. To its other end, a mass of 10 kg is attached. The tension in the string will be

\frac{g}{2}

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\frac{5 g}{2}

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\frac{10 g}{2}

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\frac{15 g}{2}

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Solution

The correct option is D $\frac{15 g}{2}$
T - 6g = 6a
10g - T = 10a $\Rightarrow T = \frac{15 g}{2}$

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A light string passes over a frictionless pulley. To one of its ends, a mass of 6 kg is attached. To its other end, a mass of 10 kg is attached. The tension in the string will be

The correct option is D 15g2T - 6g = 6a 10g - T = 10a ⇒ T=15g2

The correct option is D $\frac{15 g}{2}$
T - 6g = 6a
10g - T = 10a $\Rightarrow T = \frac{15 g}{2}$