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Physics: Units, Dimensions, Measurements and Error Analysis
Solutions of Practice Questions (MCQs) – Set 1 (Q.No. 1 – 10)
Answer 1: (d) Hint: Capacitance (C) =
, =
[AT]2
ML2T-2
or, C=
[A2T2]
ML2T-2
, = [M-1L
-2T
4A
2
].
Resistance (R) =
=
[A]
, = [ML2
T
-3A
-2].
Potential difference (V) =
, =
[AT]
or, V = [ML2T
-3A
-1] is the dimensional formula for potential difference.
Answer 2: (d). Answer 3: (b).
Answer 4: (b) Hint: =
=
Given,
∆
= ± 2% = ± 2 × 10
∆
= ± 1% = ± 1 × 10
∆
=
∆
+ 3 ∆
= 2 × 10 + 3 × 10 = 5 × 10 = 5%
Answer 5: (a). Answer 6: (d). Answer 7: (b).
Answer 8: (b) Hint: m = 1.76 kg, M = 25 m = 25 x 1.76 = 44.0 kg.
[Note: Mass of one unit has three significant figures and it is just multiplied by a pure number
(magnified). So, result should also have three significant figures].
Answer 9: (a). Answer 10: (c) Hint: Surface tension =
=
[ ]
[ ]
= [ML0
T
2
].
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