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The ratio of tensions in the string connected to the block of

Question

The ratio of tensions in the string connected to the block of mass $m_{2}$ in Fig. A.11 $(a) - (b)$ , respectively, is (friction is absent everywhere): $[m_{1} = 50 k g, m_{2} = 80 k g$ and $F = 1000 N]$

7 : 2

2 : 7

3 : 4

4 : 3

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The ratio of tensions in the string connected to the block of mass

m_{2}

in Fig. A.11

(a) - (b)

, respectively, is (friction is absent everywhere):

[m_{1} = 50 k g, m_{2} = 80 k g

and

F = 1000 N]

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Text solutionVerified

Refer above image 1.

F - T_{1} = m_{1} a ∣ T_{1} - m_{2} g = m_{2} a

a_{1} = \frac{F - m _{2} g}{m _{1} + m _{2}}

a_{1} = \frac{1000 - 800}{130}

a_{1} = \frac{200}{130}

T_{1} = F - m_{1} a

= 1000 - 50 \times \frac{200}{130}

Refer above image 2.

F + m_{1} g - T_{2} m_{1} a ∣ T_{2} - m_{2} g = m_{2} a

F + m_{1} g - m_{2} g = (m_{1} + m_{2}) a

a_{2} = \frac{1000 + 500 - 800}{130}

a_{2} = \frac{700}{130}

T_{2} = 80 (\frac{700}{130} + 10)

\frac{T _{1}}{T _{2}} = \frac{1 , 30 , 000 - 1 , 00 , 000}{56 , 000 + 1 , 04000} = \frac{3}{4}

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Question Text	The ratio of tensions in the string connected to the block of mass $m_{2}$ in Fig. A.11 $(a) - (b)$ , respectively, is (friction is absent everywhere): $[m_{1} = 50 k g, m_{2} = 80 k g$ and $F = 1000 N]$
Answer Type	Text solution:1
Upvotes	150