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Standard XII

Physics

Ropes

Two masses ...

Question

Two masses $m_{1}$ and $m_{2}$ are connected by a string passing over pulley. When set free, $m_{1}$ moves downwards by $1.4 m$ in $2 s$ . The ratio between $m_{1}$ and $m_{2}$ is

9 : 7

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11 : 9

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13 : 11

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15 : 13

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Solution

The correct option is D $15 : 13$
We can calculate $a =$ $\frac{m_{1} - m_{2}}{m_{1} + m_{2}}$
$S = \frac{1}{2} (\frac{m_{1} - m_{2}}{m_{1} + m_{2}}) g t^{2}$
$∴ \frac{m_{1} - m_{2}}{m_{1} + m_{2}} = \frac{7}{100}$
Hence by componendo dividendo:
$m_{1} : m_{2} = 107 : 93 \approx \frac{15}{13}$

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Two masses m1 and m2 are connected by a string passing over pulley. When set free, m1 moves downwards by 1.4 m in 2 s. The ratio between m1 and m2 is

The correct option is D 15:13We can calculate a= m1−m2m1+m2S=12(m1−m2m1+m2)gt2∴m1−m2m1+m2=7100Hence by componendo dividendo:m1:m2=107:93≈1513

Two masses $m_{1}$ and $m_{2}$ are connected by a string passing over pulley. When set free, $m_{1}$ moves downwards by $1.4 m$ in $2 s$ . The ratio between $m_{1}$ and $m_{2}$ is

The correct option is D $15 : 13$
We can calculate $a =$ $\frac{m_{1} - m_{2}}{m_{1} + m_{2}}$
$S = \frac{1}{2} (\frac{m_{1} - m_{2}}{m_{1} + m_{2}}) g t^{2}$
$∴ \frac{m_{1} - m_{2}}{m_{1} + m_{2}} = \frac{7}{100}$
Hence by componendo dividendo:
$m_{1} : m_{2} = 107 : 93 \approx \frac{15}{13}$