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Q.

Error in measurement of radius of a sphere is 1%, then

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a

the error in measurement in volume is 1%

b

the error in measurement of surface area is 2%

c

the error in measurement of volume is 3%

d

the error in measurement of surface area is 6%

answer is A, C.

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Detailed Solution

 V=43πR3     ΔVV=3ΔRR     ΔVV×100=3ΔRR×100=3% (3)  A=4πR2 ΔAA=2ΔRR     ΔAA×100=2ΔRR×100=2×1%=2%

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