JEE MAIN - Physics (2018 - 15th April Morning Slot - No. 18)
The relative error in the determination of the surface area of sphere is $$\alpha $$. Then the relative error in the determination of its volume is :
$${3 \over 2}\alpha $$
$${2 \over 3}\alpha $$
$${5 \over 2}\alpha $$
$$\alpha $$
Explanation
Relative error in the surface are of the sphere,
$${{\Delta S} \over S}$$ = 2 $$ \times $$ $${{\Delta r} \over r}$$ = $$ \propto $$ (given)
Relative error in volume,
$${{\Delta V} \over V}$$ = 3 $$ \times $$ $${{\Delta r} \over r}$$
= 3 $$ \times $$ $${1 \over 2}$$ $$ \times $$ $${{\Delta S} \over S}$$
= $${3 \over 2}$$ $$ \times $$ $$ \propto $$
= $${3 \over 2}$$ $$ \propto $$
$${{\Delta S} \over S}$$ = 2 $$ \times $$ $${{\Delta r} \over r}$$ = $$ \propto $$ (given)
Relative error in volume,
$${{\Delta V} \over V}$$ = 3 $$ \times $$ $${{\Delta r} \over r}$$
= 3 $$ \times $$ $${1 \over 2}$$ $$ \times $$ $${{\Delta S} \over S}$$
= $${3 \over 2}$$ $$ \times $$ $$ \propto $$
= $${3 \over 2}$$ $$ \propto $$
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