1
AIEEE 2010
+4
-1
Two conductors have the same resistance at $${0^ \circ }C$$ but their temperature coefficients of resistance are $${\alpha _1}$$ and $${\alpha _2}.$$ The respective temperature coefficients of their series and parallel combinations are nearly
A
$${{{\alpha _1} + {\alpha _2}} \over 2},\,{\alpha _1} + {\alpha _2}$$
B
$${\alpha _1} + {\alpha _2},\,{{{\alpha _1} + {\alpha _2}} \over 2}$$
C
$${\alpha _1} + {\alpha _2},\,{{{\alpha _1}{\alpha _2}} \over {{\alpha _1} + {\alpha _2}}}$$
D
$${{{\alpha _1} + {\alpha _2}} \over 2},\,{{{\alpha _1} + {\alpha _2}} \over 2}$$
2
AIEEE 2008
+4
-1
Shown in the figure below is a meter-bridge set up with null deflection in the galvanometer.

The value of the unknown resister $$R$$ is

A
$$13.75\Omega$$
B
$$220\Omega$$
C
$$110\Omega$$
D
$$55\Omega$$
3
AIEEE 2008
+4
-1
Consider a block of conducting material of resistivity $$'\rho '$$ shown in the figure. Current $$'I'$$ enters at $$'A'$$ and leaves from $$'D'$$. We apply superposition principle to find voltage $$'\Delta V'$$ developed between $$'B'$$ and $$'C'$$. The calculation is done in the following steps:
(i) Take current $$'I'$$ entering from $$'A'$$ and assume it to spread over a hemispherical surface in the block.
(ii) Calculate field $$E(r)$$ at distance $$'r'$$ from A by using Ohm's law $$E = \rho j,$$ where $$j$$ is the current per unit area at $$'r'$$.
(iii) From the $$'r'$$ dependence of $$E(r)$$, obtain the potential $$V(r)$$ at $$r$$.
(iv) Repeat (i), (ii) and (iii) for current $$'I'$$ leaving $$'D'$$ and superpose results for $$'A'$$ and $$'D'.$$

For current entering at $$A,$$ the electric field at a distance $$'r'$$ from $$A$$ is

A
$${{\rho I} \over {8\pi {r^2}}}$$
B
$${{\rho I} \over {{r^2}}}$$
C
$${{\rho I} \over {2\pi {r^2}}}$$
D
$${{\rho I} \over {4\pi {r^2}}}$$
4
AIEEE 2008
+4
-1
Consider a block of conducting material of resistivity $$'\rho '$$ shown in the figure. Current $$'I'$$ enters at $$'A'$$ and leaves from $$'D'$$. We apply superposition principle to find voltage $$'\Delta V'$$ developed between $$'B'$$ and $$'C'$$. The calculation is done in the following steps:
(i) Take current $$'I'$$ entering from $$'A'$$ and assume it to spread over a hemispherical surface in the block.
(ii) Calculate field $$E(r)$$ at distance $$'r'$$ from A by using Ohm's law $$E = \rho j,$$ where $$j$$ is the current per unit area at $$'r'$$.
(iii) From the $$'r'$$ dependence of $$E(r)$$, obtain the potential $$V(r)$$ at $$r$$.
(iv) Repeat (i), (ii) and (iii) for current $$'I'$$ leaving $$'D'$$ and superpose results for $$'A'$$ and $$'D'.$$

$$\Delta V$$ measured between $$B$$ and $$C$$ is

A
$${{\rho I} \over {\pi a}} - {{\rho I} \over {\pi \left( {a + b} \right)}}$$
B
$${{\rho I} \over a} - {{\rho I} \over {\left( {a + b} \right)}}$$
C
$${{\rho I} \over {2\pi a}} - {{\rho I} \over {2\pi \left( {a + b} \right)}}$$
D
$${{\rho I} \over {2\pi \left( {a - b} \right)}}$$
EXAM MAP
Medical
NEET