JEE MAIN - Physics (2021 - 22th July Evening Shift - No. 2)
A Copper (Cu) rod of length 25 cm and cross-sectional area 3 mm2 is joined with a similar Aluminium (Al) rod as shown in figure. Find the resistance of the combination between the ends A and B.
(Take Resistivity of Copper = 1.7 $$\times$$ 10$$-$$8 $$\Omega$$m and Resistivity of Aluminium = 2.6 $$\times$$ 10$$-$$8 $$\Omega$$m)
_22th_July_Evening_Shift_en_2_1.png)
(Take Resistivity of Copper = 1.7 $$\times$$ 10$$-$$8 $$\Omega$$m and Resistivity of Aluminium = 2.6 $$\times$$ 10$$-$$8 $$\Omega$$m)
0.0858 m$$\Omega$$
1.420 m$$\Omega$$
0.858 m$$\Omega$$
2.170 m$$\Omega$$
Explanation
$${R_{Cu}} = {{{\rho _{Cu}} \times l} \over A}$$
$${R_{Al}} = {{{\rho _{Al}} \times l} \over A}$$
$${R_{Eq}} = {{{\rho _{Cu}} \times {\rho _{Al}}} \over {{\rho _{Cu}} + {\rho _{Al}}}} \times \left( {{l \over A}} \right)$$
$$ = {{1.7 \times {{10}^{ - 8}} \times 2.6 \times {{10}^{ - 8}}} \over {(1.7 + 2.6) \times {{10}^{ - 8}}}} \times {{0.25} \over {3 \times {{10}^{ - 6}}}}$$
$$ = 0.856$$ m$$\Omega$$
$${R_{Al}} = {{{\rho _{Al}} \times l} \over A}$$
$${R_{Eq}} = {{{\rho _{Cu}} \times {\rho _{Al}}} \over {{\rho _{Cu}} + {\rho _{Al}}}} \times \left( {{l \over A}} \right)$$
$$ = {{1.7 \times {{10}^{ - 8}} \times 2.6 \times {{10}^{ - 8}}} \over {(1.7 + 2.6) \times {{10}^{ - 8}}}} \times {{0.25} \over {3 \times {{10}^{ - 6}}}}$$
$$ = 0.856$$ m$$\Omega$$
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