In the circuit shown in the figure, the emf of the battery is 9 V and its internal resistance is 15 Ω. The two identical voltmeters can be considered ideal. Let V1 and V'1 be the readings of 1st voltmeter when switch is open and closed, respectively. Similarly, V2 and V'2 the readings of 2ndvoltmeter when switch is open and closed, respectively.Column IColumn IIi.V1p.1 Vii.V'1q.2 Viii.V2r.3 Viv.V'2s.4 V Now match the given columns and select the correct option from the codes given below.

Since voltmeters are ideal, so no current flows through voltmeters. Now current I through resistances,I =950+10+20 = 945 = 15 APotential difference across 10+20 Ω=I10+20 =15 × 30=6 VThis potential will be divided equally between V1 and V2. So V1 = V2 = 3 VAfter closing the switch, current will remain same.But NowV'1 = potential difference across 10 Ω=I × 10=15 × 10 =2VV'2 potential difference across 20 Ω=I × 20=15 × 20 =4 VHence [i - r] [ii- q] [iii - r] [iv - s]

In the circuit shown in the figure, the emf of the battery is 9 V and its internal resistance is 15 Ω. The two identical voltmeters can be considered ideal. Let V1 and V'1 be the readings of 1st voltmeter when switch is open and closed, respectively. Similarly, V2 and V'2 the readings of 2ndvoltmeter when switch is open and closed, respectively.Column IColumn IIi.V1p.1 Vii.V'1q.2 Viii.V2r.3 Viv.V'2s.4 V Now match the given columns and select the correct option from the codes given below.

Since voltmeters are ideal, so no current flows through voltmeters. Now current I through resistances,I =950+10+20 = 945 = 15 APotential difference across 10+20 Ω=I10+20 =15 × 30=6 VThis potential will be divided equally between V1 and V2. So V1 = V2 = 3 VAfter closing the switch, current will remain same.But NowV'1 = potential difference across 10 Ω=I × 10=15 × 10 =2VV'2 potential difference across 20 Ω=I × 20=15 × 20 =4 VHence [i - r] [ii- q] [iii - r] [iv - s]