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Byju's Answer

Standard XII

Physics

Tension in a String

In the figure...

Question

In the figure shown, the $12 k g$ body pulled by a string with an acceleration $a = 2.2 m / s^{2}$ . The tension $T_{1}$ and $T_{2}$ will be respectively
(Take $g = 9.8 m / s^{2}$ )

200 N, 80 N

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240 N, 90 N

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240 N, 96 N

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260 N, 96 N

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Solution

The correct option is C $240 N, 96 N$
Considering FBD of combined mass of $20 k g$ ,

$T_{1} - m g = m a = 20 (9.8 + 2.2) = 240 N$
Considering FBD of $8 k g$ mass,

$T_{2} - m g = m a = m (a + g) = 8 \times (2.2 + 9.8) = 96 N$

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A body of mass 10kg is hanging from another body of mass 15kg. The combination is pulled up by a string with an acceleration of . $1.2 m^{- 2}$ Find the tensions (in N) at A and B points as shown in figure $(T a k e g = 9.8 m / s^{2})$ .

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In the figure shown, the 12 kg body pulled by a string with an acceleration a=2.2 m/s2. The tension T1 and T2 will be respectively (Take g=9.8 m/s2)

The correct option is C 240 N, 96 NConsidering FBD of combined mass of 20 kg, T1−mg=ma=20(9.8+2.2)=240 N Considering FBD of 8 kg mass, T2−mg=ma=m(a+g)=8×(2.2+9.8)=96 N

In the figure shown, the $12 k g$ body pulled by a string with an acceleration $a = 2.2 m / s^{2}$ . The tension $T_{1}$ and $T_{2}$ will be respectively
(Take $g = 9.8 m / s^{2}$ )

The correct option is C $240 N, 96 N$
Considering FBD of combined mass of $20 k g$ ,

$T_{1} - m g = m a = 20 (9.8 + 2.2) = 240 N$
Considering FBD of $8 k g$ mass,

$T_{2} - m g = m a = m (a + g) = 8 \times (2.2 + 9.8) = 96 N$