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Question

Three rivers R1,R2 and R3 merge together to form a river, R4. The cross sectional areas of the three rivers R1,R2 and R3 are in the ratio 1:2:3 and speed of water flowing in these are in the ratio 1:12:13. Assuming streamline flow, if R4 has cross sectional area equal to that of R1, the ratio of speed of water in R4 to that in R1 is

A
4:1
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B
3:1
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C
2:1
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D
1:1
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Solution

The correct option is B 3:1
Discharge=velocity×cross section area
R1:R2:R3=area=1:2:3
R1=1x,R2=2x,R3=3x are the cross section areas
and their respective velocities are 1:1/2:1/3=6:3:2
R1=6y,R2=3y,R3=2y
R1 discharge =1x×6y=6xy
R2.=6xy
R3.=6xy
R4 discharge =R1+R2+R3=18xy
R4 has he same cross section as R1=1x
so, R4 has velocity =18y
R4:R3(velocity ratio)=18y:6y
which is3:1

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