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An open-ended U-tube of uniform cross-sectional area contains water (density \mathrm{1.0 \mathrm{gram} / centimeter ^3} ) standing initially 20 centimeters from the bottom in each arm. An immiscible liquid of density \mathrm{4.0 \, \, \mathrm{grams} /}centimeter \mathrm{{ }^3} is added to one arm until a layer 5 centimeters high forms, as shown in the figure above. What is the ratio \mathrm{h_2 / h_1} of the heights of the liquid in the two arms?

Option: 1

3 / 1


Option: 2

5 / 2


Option: 3

2 / 1


Option: 4

3 / 2


Answers (1)

best_answer

\mathrm{\begin{aligned} & P_A=P_B \\\\ & \Rightarrow 5 \times 4 \times g+x \times 1 \times g \\\\ & =(40-x) \times 1 \times g \\\\ & \Rightarrow x=10 \\\\ & \text { Now, } h_1=x+5=15 \mathrm{~cm} \\\\ & h_2=40-x=30 \mathrm{~cm} \\\\ & h_2 / h_1=2 \end{aligned}}

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Sanket Gandhi

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