Two ideal springs with light pans are hanging vertically as shown. 1 kg mass is placed in pan A, the descent of pan B in equilibrium is x1. When the same mass is instead placed in pan B the descent of pan B in equilibrium is x2. Then x2/x1 :

x1=mgk1=mg10x2=mgk1k2k1+k2=mg200/30

Two ideal springs with light pans are hanging vertically as shown. 1 kg mass is placed in pan A, the descent of pan B in equilibrium is x1. When the same mass is instead placed in pan B the descent of pan B in equilibrium is x2. Then x2/x1 :

x1=mgk1=mg10x2=mgk1k2k1+k2=mg200/30

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Two ideal springs with light pans are hanging vertically as shown. 1 kg mass is placed in pan A, the descent of pan B in equilibrium is x₁. When the same mass is instead placed in pan B the descent of pan B in equilibrium is x₂. Then x₂/x₁:

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$\frac{2}{3}$

$\frac{1}{2}$

None

$\frac{3}{2}$

answer is A.

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$\begin{array}{l} x_{1} = \frac{mg}{k_{1}} = \frac{mg}{10} \\ x_{2} = \frac{mg}{(\frac{k_{1} k_{2}}{k_{1} + k_{2}})} = \frac{mg}{200 / 30} \end{array}$

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Two ideal springs with light pans are hanging vertically as shown. 1 kg mass is placed in pan A, the descent of pan B in equilibrium is x1. When the same mass is instead placed in pan B the descent of pan B in equilibrium is x2. Then x2/x1 :

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Two ideal springs with light pans are hanging vertically as shown. 1 kg mass is placed in pan A, the descent of pan B in equilibrium is x₁. When the same mass is instead placed in pan B the descent of pan B in equilibrium is x₂. Then x₂/x₁: