The figure given below shows a circuit when resistances in the two arms of the meter bridge are \(5~\Omega\) and \(R\), respectively. When the resistance \(R\) is shunted with equal resistance, the new balance point is at \(1.6l_1\). The resistance \(R\) is: 
           
1. \(10~\Omega\)
2. \(15~\Omega\)
3. \(20~\Omega\)
4. \(25~\Omega\)

A wire of resistance \(4~\Omega\) is stretched to twice its original length. The resistance of a stretched wire would be:
1. \(4~\Omega\)
2. \(8~\Omega\)
3. \(16~\Omega\)
4. \(2~\Omega\)

If the voltage across a bulb rated \((220~\text{V}\text-100~\text{W})\) drops by \(2.5\%\) of its rated value, the percentage of the rated value by which the power would decrease is:
1. \(20\%\)
2. \(2.5\%\)
3. \(5\%\)
4. \(10\%\)

A ring is made of a wire having a resistance of \(R_0=12~\Omega.\). Find points \(\mathrm{A}\) and \(\mathrm{B}\), as shown in the figure, at which a current-carrying conductor should be connected so that the resistance \(R\) of the subcircuit between these points equals \(\frac{8}{3}~\Omega\)$\mathrm{?}$
      
1. \(\frac{l_1}{l_2} = \frac{5}{8}\)
2. \(\frac{l_1}{l_2} = \frac{1}{3}\)
3. \(\frac{l_1}{l_2} = \frac{3}{8}\)
4. \(\frac{l_1}{l_2} = \frac{1}{2}\)

If power dissipated in the \(9~\Omega\) resistor in the circuit shown is \(36\) W, the potential difference across the \(2~\Omega\) resistor will be:

A current of \(2~\text{A}\) flows through a \(2~\Omega\) resistor when connected across a battery. The same battery supplies a current of \(0.5~\text{A}\) when connected across a \(9~\Omega\) resistor. The internal resistance of the battery is:
1. \(\frac{1}{3}~\Omega\)
2. \(\frac{1}{4}~\Omega\)
3. \(1~\Omega\)
4. \(0.5~\Omega\)

A wire of resistance \(12~ \Omega \text{m}^{-1}\)${\mathrm{}}^{}$ is bent to form a complete circle of radius \(10~\text{cm}\). The resistance between its two diametrically opposite points, \(A\) and \(B\) as shown in the figure, is:
        
1. \(0.6\pi~\Omega\)$$
2. \(3\pi ~\Omega\)$$
3. \(61 \pi~ \Omega\)$$
4. \(6\pi~\Omega\)$$