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Answer 1

$\blue{\bold{\underline{\underline{Answer:}}}}$

$\green{\tt{\therefore{\frac{m_{1}}{m_{2}}=\frac{7}{5}}}}$

$\orange{\bold{\underline{\underline{Step-by-step\:explanation:}}}}$

$\green{\underline \bold{Given :}} \\ \tt: \implies Acceleration \: of \: system = \frac{g}{6} \\ \\ \red{\underline \bold{To \: Find :}} \\ \tt: \implies \frac{ m_{1} }{m_{2}} =?$

• According to given question :

$\tt \circ \: Let \: mases \: be \: m_{1} \: and \: m_{2} \\ \\ \tt \circ \: m_{1} > m_{2} \\ \\ \bold{As \: we \: know \: that} \\ \tt: \implies m_{1}g - T = m_{1}a \\ \\ \tt: \implies m_{1}g - m_{1} \times \frac{g}{6} = T \\ \\ \tt: \implies T = \frac{5 m_{1}g }{6} - - - - - (1) \\ \\ \bold{Again :} \\ \tt: \implies T- m_{2}g = m_{2}a \\ \\ \tt: \implies \frac{5 m_{1}g}{6} = m_{2} \times \frac{g}{6} + m_{2}g \\ \\ \tt: \implies \frac{5 m_{1}g }{6} = \frac{7 m_{2}g }{6} \\ \\ \tt: \implies \frac{ m_{1}}{ m_{2} } = \frac{7 \times 6}{6 \times 5} \\ \\ \green{\tt: \implies \frac{ m_{1}}{ m_{2} } = \frac{7}{5} }$

Answer 2

GIVEN:

• Two masses m1 & m2 are connected by a light string passing over a pulley
• Acceleration of the system g/6
• m1 > m2

TO FIND:

• m1 : m2

SOLUTION:

We know that

→ m1g - T = m1a

Substituting a = g/6. ( °.° Given )

→ m1g - m1(g/6) = T

→ 5m1g/6 = T

Assuming as eq 1

________________________

→ T - m2g = m2a

→ T - m2g = m2(g/6)

→ T = 7m2g/6

Assuming as eq 2

Now, Eq 2 ÷ 1

♦ T/T = [7m2g/6]/[5m1g/6]

♦ 1 = 7m1/5m2

♦ m2/m1 = 5/7

♦ m1 : m2 = 7 : 5. ( Option c )

Hence, option c is your answer.