Two bodies of masses 1 kg and 2 kg are connected by a metal wire as shown in figure. A force of 10 N is applied on the body of mass 2 kg. The breaking stress of metal wire is 2 × 109 Nm -2. What should be the radius of wire used, if it is not to break

Question

Two bodies of masses 1 kg and 2 kg are connected by a metal wire as shown in figure. A force of 10 N is applied on the body of mass 2 kg. The breaking stress of metal wire is 2 × 109 Nm -2. What should be the radius of wire used, if it is not to break (A) 4 × 10-4 m (B) 5 × 10-4 (C) 0.23 × 10-4 (D) 5.2 × 10-4 m

Tardigrade Admin · Accepted Answer

T=(m2 F/m1+m2)=(1 × 10/3)=(10/3) N (F/A)=2 × 109 Nm -2 (10/3 × π r2)=2 × 109 ⇒ r2=(10-8/6 π) r=(10-4/√6 π)=0.23 × 10-4 m

Q. Two bodies of masses $1 k g$ and $2 k g$ are connected by a metal wire as shown in figure. A force of $10 N$ is applied on the body of mass $2 k g$ . The breaking stress of metal wire is $2 \times 1 0^{9} N m^{- 2}$ . What should be the radius of wire used, if it is not to break

$4 \times 1 0^{- 4} m$

$5 \times 1 0^{- 4}$

$0.23 \times 1 0^{- 4}$

$5.2 \times 1 0^{- 4} m$

$T = \frac{m _{2} F}{m _{1} + m _{2}} = \frac{1 \times 10}{3} = \frac{10}{3} N$
$\frac{F}{A} = 2 \times 1 0^{9} N m^{- 2}$
$\frac{10}{3 \times π r ^{2}} = 2 \times 1 0^{9}$
$\Rightarrow r^{2} = \frac{1 0 ^{- 8}}{6 π}$
$r = \frac{1 0 ^{- 4}}{6 π} = 0.23 \times 1 0^{- 4} m$

Q. Two bodies of masses 1kg and 2kg are connected by a metal wire as shown in figure. A force of 10N is applied on the body of mass 2kg. The breaking stress of metal wire is 2×109Nm−2. What should be the radius of wire used, if it is not to break

4×10−4m

5×10−4

0.23×10−4

5.2×10−4m