Question

A force of $10 ^ { 3 }$ newton stretches the length of a hanging wire by 1 millimetre. The force required to stretch a wire of same material and length but having four times the diameter by 1 millimetre is

• A. $4 \times 10 ^ { 3 }$N
• B. $16 \times 10 ^ { 3 }$N
• C. $\frac { 1 } { 4 } \times 10 ^ { 3 }$N
• D. $\frac { 1 } { 16 } \times 10 ^ { 3 }$N

Answer 1

Answer: (B)

$16 \times 10 ^ { 3 }$N

Answer 2

$F = Y \times A \times \frac { l } { L }$⇒$F \propto r ^ { 2 }$(Y,l and L are constant)

If diameter is made four times then force required will be 16 times. i.e. 16 × 10³ N