Q1: Which of the following combinations has the dimension of electrical resistance (∈0 is the permittivity of vacuum and μ0 is the permeability of vacuum)?
(a)
(b)
(c)
(d)
Ans: (c)
According to Coulomb's law
Force between two parallel current carrying wires,
From Ohm's law,V = IR
By comparing both sides we get,- a + b = 1
- 3a + b = 2
By solving we get,
Q2: In the formula X = 5YZ2 , X and Z have dimensions of capacitance and magnetic field, respectively. What are the dimensions of Y in SI units?
(a) [M–3L–2T8A4]
(b) [M–2L–2T6A3]
(c) [M–1L–2T4A2]
(d) [M–2L0 T–4A–2]
Ans: (a)
Given,X = 5YZ2
∴ [Y] = [M–3L–2T8A4]
Q3: The area of a square is 5.29 cm2. The area of 7 such squares taking into account the significant figures is:-
(a) 37.0 cm2
(b) 37 cm2
(c) 37.030 cm2
(d) 37.03 cm2
Ans: (a)
The area of one square is 5.29 cm².
To find the area of 7 such squares, we can simply multiply the area of one square by 7:
7 x 5.29 cm² = 37.03 cm²
Since the given area has three significant figures, we need to round our answer to three significant figures as well.
Therefore, the area of 7 such squares, taking into account the significant figures, is 37.0 cm².
Q4: In the density measurement of a cube, the mass and edge length are measured as (10.00 ± 0.10) kg and (0.10 ± 0.01) m, respectively. The error in the measurement of density is :
(a) 0.01 kg/m3
(b) 0.10 kg/m3
(c) 0.31 kg/m3
(d) 0.07 kg/m3
Ans: (c)
Mass (m) = (10.00 ± 0.10) kg
Edge length (l) = (0.10 ± 0.01) m
Volume of the cube (V) = l3
Q5: In a simple pendulum experiment for determination of acceleration due to gravity (g), time taken for 20 oscillations is measured by using a watch of 1 second least count. The mean value of time taken comes out to be 30 s. The length of pendulum is measured by using a meter scale of least count 1 mm and the value obtained is 55.0 cm. The percentage error in the determination of g is close to :-
(a) 0.2%
(b) 3.5%
(c) 0.7%
(d) 6.8%
Ans: (d)
Time period of a pendulum
Fractional change
∴ Maximum possible percentage error,
Error in time period(dT) = least count of time = 1 second
and T = 30 second
Error in length(dl) = least count of length = 1 mm
and l = 55.0 cm
Q6: If surface tension (S), Moment of inertia (I) and Planck's constant (h), were to be taken as the fundamental units, the dimensional formula for linear momentum would be :-
(a) S1/2I1/2h0
(b) S3/2I1/2h0
(c) S1/2I1/2h-1
(d) S1/2I3/2h-1
Ans: (a)
We know,
surface tension
Moment of inertia (I) = mr2
∴ [I] = [ML2]
Planck's constant (h) = E/f = Et
∴ [h] = [ML2T−1]
Also linear momentum (p) = mv = [M LT−1]
Now we have to express p in terms of s, I and h.
∴ Let, [P] = [Sa Ib hc]
By comparing the dimensions of both sides, we get
a + b + c = 1 .........(1)
2b +2c = 1 ..............(2)
- 2a - c = -1 ...................(3)
By solving those three equations we get,
a = 1/2
b = 1/2
c = 0
∴ linear momentum [p] = [S1/2I1/2h0]
Q7: In SI units, the dimesions of
(a) A–1 TML3
(b) A2T3M–1L–2
(c) AT–3ML3/2
(d) AT2M–1L–1
Ans: (b)
Q8: Let ℓ, r, C and V represent inductance, resistance, capacitance and voltage, respectively. The dimension of in SI units will be:
(a) [A–1]
(b) [LTA]
(c) [LA–2]
(d) [LT2]
Ans: (a)
Q9: The least count of the main scale of a screw gauge is 1 mm. The minimum number of divisions on its circular scale required to measure 5 μm diameter of a wire is :
(a) 500
(b) 100
(c) 200
(d) 50
Ans: (c)
N = 200
Q10: If speed (V), acceleration (A) and force (F) are considered as fundamental units, the dimension of Young,s modulus will be:
(a) V−2A2F2
(b) V−4A−2F
(c) V−4A2F
(d) V−2A2F−2
Ans: (c)
We know,
For dimensional balance, the dimension on both sides should be same.
So, z = 1
x + y + z = -1
⇒ x + y = -2 ........(1)
and -x -2y - 2z = -2
⇒ x + 2y = 0 ...........(2)
By solving those two equations we get,
x = -4 and y = 2
∴ [Y] = V−4A2F1
Q11: The force of interaction between two atoms is given by where x is the distance, k is the Boltzmann constant and T is temperature and α and β are two constants. The dimension of β is :
(a) M2L2T−2
(b) M2LT−4
(c) MLT−4
(d) M0L2LT−4
Ans: (b)
Q12: The diameter and height of a cylinder are measured by a meter scale to be 12.6 ± 0.1 cm and 34.2 ± 0.1 cm, respectively. What will be the value of its volume in appropriate significant figures ?
(a) 4264.4 ± 81.0 cm3
(b) 4264 ± 81 cm3
(c) 4300 ± 80 cm3
(d) 4260 ± 80 cm3
Ans: (d)
Volume of cylinder(V) = πr2h
∴ ΔV = 0.0188 × 4260 = 80
Q13: The density of a material in SI units is 128 kg m–3 . In certain units in which the unit of length is 25 cm and the unit of mass is 50 g, the numerical value of density of the material is -
(a) 40
(b) 640
(c) 16
(d) 410
Ans: (a)
Here given that
density of a material in SI units is 128 kg m–3
And a new unit system is introduced where 1 unit of length = 25 cm and 1 unit of mass = 50 g
You should know that, physical quantity is same in any unit system. And to calculate a physical quantity yo should know two things
(1) numerical value of the physical quantity (n)
(2) unit of the physical quantity (u)
And n × u = constant in any unit system.
Here in SI unit system,
n1 = 128
u1 = kg/m3
And in new unit system,
n2 = ?
u2 = 50gm/(25cm)3
As n1u1 = n2u2
∴ 128 × (kg/m3) = n2 × 50gm/(25cm)3
Q14: Expression for time in terms of G(universal gravitional constant), h (Planck constant) and c (speed of light) is proportional to :
(a)
(b)
(c)
(d)
Ans: (c)
Let t ∝ Gx hy cz
∴ [t] = [G]x [h]y [c]z . . . . . (1)
We know,
[G] = [M−1 L3 T−2]
Also,
E = hf
From equation (1) we get,
By comparing the power of M, L, T
− x + y = 0
⇒ x = y
3x + 2y + z = 0
⇒ 5x + z = 0 . . . . . (2)
− 2x − y − z = 1
⇒ − 3x − z = 1 . . . . (3)
By solving (2) and (3), we get,
Q15: The pitch and the number of divisions, on the circular scale, for a given screw gauge are 0.5 mm and 100 respectively. When the screw gauge is fully tightened without any object, the zero of its circular scale lies 3 divisions below the mean line.
The readings of the main scale and the circular scale, for a thin sheet, are 5.5 mm and 48 respectively, the thickness of this sheet is:
(a) 5.755 mm
(b) 5.950 mm
(c) 5.725 mm
(d) 5.740 mm
Ans: (c)
We know,
= 0.5 × 10−2 mm
Reading = MSR + CSR − positive error
Given, Main scale reading (MSR) = 5.5 mm
Circular scale reading (CSR)
= 48 × 0.5 × 10−2 mm
= 0.24
As zero of its circular scale lines 3 division below the mean line, it means error is position error.
∴ positive error
= 3 × 0.5 × 10−2 mm
= 0.015 mm
∴ Reading = 5.5 + 0.24 − 0.015
= 5.725 mm
Q16: A copper wire is stretched to make it 0.5% longer. The percentage change in its electrical resistance if its volume remains unchanged is:
(a) 2.0 %
(b) 2.5 %
(c) 1.0 %
(d) 0.5 %
Ans: (c)
We know,
and Volume (V) = Al
= 2 × 0.5
= 1%