The heat developed in a wire is given by Joule’s law: \[ H = I^2 R t \] where \( H \) is the heat developed, \( I \) is the current, \( R \) is the resistance, and \( t \) is the time. The resistance of a wire is given by: \[ R = \rho \frac{L}{A} \] where \( \rho \) is the resistivity, \( L \) is the length, and \( A \) is the cross-sectional area of the wire. For a wire with radius \( r \), the cross-sectional area is \( A = \pi r^2 \). To double the heat developed, we need to double the resistance. Since resistance is directly proportional to the length and inversely proportional to the cross-sectional area, doubling both the length and radius of the wire will double the resistance, thus doubling the heat developed.
Hence, the correct answer is (b).

A capacitor of capacitance 5 µF is connected as shown in the figure. The internal resistance of the cell is 0.5 Ω. The amount of charge on the capacitor plate is: 
The resistance \( R \) of a conductor varies with temperature \( t \) as shown in the figure. If the variation is represented by \( R_t = R_0 \left( 1 + \alpha t + \beta t^2 \right) \), then: 