A) zero
B) 2R
C) 8R
D) 16R
Correct Answer: D
Solution :
Key Idea: Volume with remain constant on stretching the wire. For a wire of length ?, area A and specific resistance \[\rho \], resistance (R) is given by \[\therefore \] ??.(1) Also volume = length \[V=E-\left( \frac{E}{R+r} \right)r\] area which remains constant on stretching the wire, hence \[E=2V,\,r=0.1\Omega ,R=3.9\Omega \] If \[V=2-\left( \frac{2}{3.9+0.1} \right)\times 0.1\] and \[V=1.95V\] are radii of wire then \[{{m}_{1}}{{u}_{1}}={{m}_{2}}{{v}_{2}}\] Using Eq. (1), we have \[{{u}_{1}}=1m/s=0.05kg,{{v}_{2}}=30m/s\] \[\Rightarrow \] \[{{m}_{1}}\times 1=0.05\times 30\] \[\Rightarrow \]
You need to login to perform this action.
You will be redirected in
3 sec
