The resistance of a conductor at a temperature \( T \) is given by: \[ R_T = R_0 \left( 1 + \alpha (T - T_0) \right), \] where: \( R_T = 1.25R_0 \) (final resistance is 25% greater than the initial resistance), \( R_0 \) is the resistance at \( T_0 = 27^\circ \text{C} \), \( \alpha = 2.0 \times 10^{-4} \, \text{C}^{-1} \) (temperature coefficient of resistance). Substitute \( R_T = 1.25R_0 \) into the equation: \[ 1.25R_0 = R_0 \left( 1 + \alpha (T - T_0) \right). \] Simplify: \[ 1.25 = 1 + \alpha (T - 27). \] Rearrange: \[ \alpha (T - 27) = 0.25. \] Substitute \( \alpha = 2.0 \times 10^{-4} \): \[ (2.0 \times 10^{-4}) (T - 27) = 0.25. \] Solve for \( T \): \[ T - 27 = \frac{0.25}{2.0 \times 10^{-4}} = 1250. \] \[ T = 27 + 1250 = 1277^\circ \text{C}. \] Answer: The temperature is \( 1277^\circ \text{C} \).
Two cells of emf 1V and 2V and internal resistance 2 \( \Omega \) and 1 \( \Omega \), respectively, are connected in series with an external resistance of 6 \( \Omega \). The total current in the circuit is \( I_1 \). Now the same two cells in parallel configuration are connected to the same external resistance. In this case, the total current drawn is \( I_2 \). The value of \( \left( \frac{I_1}{I_2} \right) \) is \( \frac{x}{3} \). The value of x is 1cm.
Current passing through a wire as function of time is given as $I(t)=0.02 \mathrm{t}+0.01 \mathrm{~A}$. The charge that will flow through the wire from $t=1 \mathrm{~s}$ to $\mathrm{t}=2 \mathrm{~s}$ is:
The correct IUPAC name of \([ \text{Pt}(\text{NH}_3)_2\text{Cl}_2 ]^{2+} \) is: