MCQGeeks
0 : 0 : 1
CBSE
JEE
NTSE
NEET
English
UK Quiz
Quiz
Driving Test
Practice
Diagrams
Games
NEET
Physics NEET MCQ
Quiz 6
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
26
27
28
29
30
Q.1
If dimensional formula of physical quantity id Ma LbTc. If a = 1, b = -1, c = -2 then the physical quantity is ... ...
0%
a) Pressure
0%
b) Force
0%
c) Angular momentum
0%
d) Energy
Explanation
Pressure = Force /Area Answer : (a)
Q.2
Dimensional formula of root means square of velocity is ...
0%
a) M0L1T-1
0%
b) M0L-1T-1
0%
c) M0L0T-1
0%
d) M0L2T-1
Explanation
Answer : (a)
Q.3
SI unit of pressure is ... ...
0%
a) pascal
0%
c) cm of Hg
0%
d) atmosphere
0%
b) dyne/cm²
Explanation
Answer : (a)
Q.4
Dimensions of coefficient of thermal conductivity is
0%
a) ML2T-2K-1
0%
b) MLT-3K-1
0%
c) MLT-2K-1
0%
d) MLT-3K
Explanation
Heat conducted by rod is given by formula here k is Coefficient of conductivity T1 and T2 are temperature A: Area norml to the heat flow d= length of rod Since equation is dimensionality valid Answer:(b)
Q.5
The damping force on a body is proportional to the velocity of the body. In this position the proportionality constant has dimensions of ..
0%
a) M2T-1
0%
b) MLT-2
0%
c) kg/ sec
0%
d) It is the dimensionless constant
Explanation
F = k V k = F/V Thus dimension of k = [MLT-2] / [LT-1] Dimension of k = MT-1= kg/sec Answer: (c)
Q.6
Curie is a unit of ..
0%
a) Energy of γ rays
0%
b) Half life
0%
c) Radioactivity
0%
d) Intensity of γ rays
Explanation
Answer: (c)
Q.7
Oersted is the unit of
0%
a) Dip
0%
b) Magnetic intensity
0%
c) Magnetic moment
0%
d) pole strength
Explanation
Answer: (b)
Q.8
Ampere-hour is unit of
0%
a) Quantity of electricity
0%
b) Strength of electric current
0%
c) power
0%
d) energy
Explanation
Qty of charge = Current × time Answer:(a)
Q.9
Dimension of coefficient of viscosity re
0%
a) ML2T-2
0%
b) ML2T-1
0%
c) ML-1T-1
0%
d) MLT
Explanation
Viscus Force F = η A (dV/dx) [MLT-2 = η[L2] [ T-1] η = ML-1T-1 Answer: (c)
Q.10
Unit of Stefan's constant is
0%
a) Js-1
0%
b) Jm-2s-1K-4
0%
c) Jm-2
0%
d) Js
Explanation
Stefan's law: Radiant energy per emitted by unit area of surface in unit second is proportional to the fourth power of absolute temperature of the body E=σT4 --(1) Here T is temperature Now E = Energy / (Area-time) Unit of E = Jm-2s-1 Thus σ = E/T4 unit of σ = Jm-2s-1 / K4 Unit of σ = Jm-2s-1 K-4 Answer: (b)
Q.11
Dimensional formula for Boltzmann constant K ..
0%
a) [ML2T-2θ-1 mole-1]
0%
b) [ML2T-2θ-1 ]
0%
c) [M2LT-2θ]
0%
d) [ML0T-2θ-1 mole-1]
Explanation
According to Ideal gas equation PV = nRT R = PV/nT PV have dimension of work thus dimension of R is ML2T-2θ-1mole-1 Now Boltzman constant = R/NA Here NA is Avogadro's number Thus dimension of Boltzmann constant = [ML2T-2θ-1 mole-1] Answer: (a)
Q.12
the dimensional formula for bulk modulus of elasticity is .. [MNR1986]
0%
a) ML-2T-2
0%
b) ML-3T-2
0%
c) ML2T-2
0%
d) ML-1T-2
Explanation
Bulk modulus = (-ΔP V) / ΔV As V/ΔV is dimension less Dimension of Bulk modulus = Dimension of pressure pressure = F/A Dimension of pressure = [ML-1T-2] Answer:(d)
Q.13
Which of the following is different from others
0%
a) Plank's constant
0%
b) Coefficient of viscosity
0%
c) Force constant
0%
d) Poission's ratio
Explanation
Poission's ratio don't have dimensions Answer: (d)
Q.14
In Ohm's experiment, the value of an unknown resistance were found to be 4.12Ω, 4.08Ω, 4.22Ω and 4.14Ω. Calculate absolute error, relative error and percentage error. in these measurement. ... .. [ Guj. T.B. ]
0%
a) 4.14, 0.04, 0.096%
0%
b) 0.04, 0.0096, 0.96%
0%
c) 0.04, 0.96, 96%
0%
d) 4.14, 0.4, 0.0096%
Explanation
Average resistance = (4.12+4.08+4.22+4.14)/4 = 4.14Ω Absolute error of each observation ΔR1 = 4.14 - 4.12 = 0.02 Ω ΔR2 = 4.14 - 4.08 = 0.06 Ω ΔR3 = 4.14 - 4.22 = -0.08 Ω ΔR4 = 4.14 - 4.14 = 0.00 Ω Average absolute error, considering only magnitude Average absolute error = (0.2 + 0.6 + 0.08)/4 = 0.04Ω Resistance R = (4.14 ±0.04)Ω Relative error = Absolute error/ average value = 0.04/4.14 = 0.0096 Percentage error = Relative error ×100 Percentage error = 0.0096 ×100 = 0.96% Answer : (b)
Q.15
The length breadth and thickness of metal sheet are 4.234 m., 1.005 m and 2.01 cm respectively. Calculate the total area and volume of the sheet to the correct number of significant figures.
0%
a) 8.7216 m2, 0.855 m3
0%
b) 8.72 m2, 0.086 m3
0%
c) 8.721 m2, 0.0855 m3
0%
d) 8.7205 m2, 0.8553 m3
Explanation
length = 4.234 m breadth b = 1.005 m height = h = 2.01 cm = 0.0201 m Total area = 2 [ (l × b) + (b × h) + (h × l) ] Total area = 8.7209478 m3 Since minimum significant figure are three in data thus answer should have only three significant figure Correct answer = 8.72 m2 Volume = l × b × h Volume = 4.234 ×1.005 × 0.0201 = 0.0855289 Since minimum significant figure are three in data thus answer should have only three significant figure V = 0.086 m3 Answer : (b)
Q.16
The electric force between two electric charges is given by by formula Here r is the distance between changes q1 and qGive the unit and dimensional formula of ε0
0%
a) N-1C2m-2, M-1 L-3T4 A2
0%
b) N2C-1m2, M-1 L3T4 A-2
0%
c) N1C-2m3, M1 L3T-4 A-2
0%
d) N1C-2m3, M1 L-3T-4 A2
Explanation
Answer : (a)
Q.17
Check dimensional validity of the following equations A) Pressure P = h ρ g All the terms have usual meaning
0%
a) A and B both are dimensionally valid
0%
b) A and B are dimensionally invalid
0%
c) A is dimensionally invalid and B is dimensionally valid
0%
d) A is dimensionally valid and B is dimensionally invalid
Explanation
A) Pressure P= h ρ g Dimensions of LHS = [P] = M1 L-1 T-2 Dimension of RHS = [h ρ g ] = [ M1L3T0] [M0 L1 T-2] [ M0 L1 T0] Dimensions of RHS = M1 L-1 T-2 Since dimensions of LHS = RHS A) is dimensionally valid For B) Dimension of LHS = [s] = [M0 L1 T0] Now RHS have two terms first v0 t Dimension of first RHS term = [ M0 L1 T-1] [ M0L0 T1] Dimension of first term = [M0 L1 T0] Dimensions of second terms = [at]2 = [a]2 [t]2 = [M0L1 T-2]2 [M0 L0 T1]2 = [M0L2T-2] Thus second terms of RHS is not equal to LHS thus B is dimensionally in-valid Answer : (d)
Q.18
Check dimensional validity of the following equations
0%
a) A and B are dimesionally valid
0%
b) A and B are dimensionally invalid
0%
c) A is dimensionally valid nad B is dimensionally invalid
0%
d) A is dimensionally invalid, B is dimensionally valid
Explanation
First equation A) Dimensions of LHS = [Fs] = [ M1L1 T-2] [ M0L1 T1] = [M1 L2 T-2 ] Dimensions of RHS [mv02] = [mv2 ] = [M1 L0 T0] [ M0L1T-2]2 =M1L2T-2 Thus LHS = RHS A) dimensionally valid Second equation B) Dimensions of LHS = [F] = [M1L1T-2] Dimensions of RHS = Dimensions of LHS ≠ RHS B) is dimensionally invalid Answer : (c)
Q.19
Above equation is dimensionally valid equation. Obtain dimensional formula of a, b c. Here v is velocity v0 is initial velocity, t is time
0%
a) LT-2, L, T
0%
b) L1, T L2 T1
0%
c) L2T1, L1T2, L
0%
d) l, LT, T2
Explanation
Addition and subtraction can be carried out with physical quantities having same dimensions only there is term t + c , c will have same dimensions of t or [c] = [M0 L0 T1] Now terms b/(t+c) should have dimensions of velocity [M0L1 T-1 ] = [b] / [M0 L0 T1] On simplification we get [b] = M1L1T0 Dimension of at should havae dimension of v thus [at] = [v] [a] [ M0 L0 T1 ] = [ M0 L1 T-1] on simplification we get [a] = M0 L1 T-2 Answer : (a)
Q.20
An object is falling freely under the gravitational force. Its velocity after travelling a distance h is v. if speed depends upon gravitational acceleration g and distance. hnece v ∝ .... [ Guj. T.B. ]
0%
a) √(gh)
0%
b) gh
0%
c) g√h
0%
d) h√g
Explanation
Suppose velocity of object v ∝ ga hb v = kga hb let k be dimension less constant, equation is dimensionally valid [v] = [g]a [ h]b [M0L1T-1 ] = [M0L1T-2]b [ M0L1T0]b0L1T-1 ] = [M0La+b T-2b Comparing the base we get a+b = 1 and -2a = -1 Thus a = 1/2 and b = 1/2 Thus g ∝ √(gh) Answer : (a)
Q.21
A gas bubble formed due to an explosion under water oscillates with a period T proportional to Pa ρbEc. Where P is the static pressure, ρ is density of water and E is the total energy of the explosion. Find value of a , b and c ... ... [ Guj T.B]
0%
a) 1, 1, 1
0%
b) 1/4, 1/2, 1/3
0%
c) -5/6, 1/2, 1/3
0%
d) 1/3, 1/2, -5/6
Explanation
T ∝ k Pa ρbEc Let proportionality constant be dimension less [M0 L0T1] = [ M1 L-1 T-2]a [M1 L-3T0]b [M1L2 T-2]c [M0 L0T1] = Ma+b+c L-a-3b+2c T-2a-2c comparing indices of M a+b+c = 0 Comparing indices of L -a-3b+2c = 0 Comparing indices of T -2a-2c = 1 On simplification of above equation we get a= -5/6, b = 1/2 and c = 1/3 Answer : (c )
Q.22
If the velocity of light, acceleration due to gravity nad normal pressure are chosen as fundamental units, find the unit of mass, length and time. Given velocity of light c = 3×108, g = 10ms-2 and normal atmospheric pressure P = 105 N/m2
0%
a) 8.1 ×1035 kg, 9 ×1015 m, 3 ×107s
0%
b) 9.12 ×1015 kg, 3 ×107 m, 8 ×1035s
0%
c) 3.02 ×107 kg, 8.1 ×1035 m, 9 ×1015s
0%
d) 1035 kg, 9 ×1015 m, 3 ×107s
Explanation
Given [c] = L1T-1 = 3 × 108 ms-1 ... .... eq(1) [g] = L1T-2 = 10 ms-2 ... .... eq(2) [P] = M1 L-1T-2 = 105 Nm-2 ... ... eq(3) Dividing eq(1) by eq(2) Substituting value of T in eq(1) L1T-1 = 3 × 108 L = 3 ×108 ×3×107 = 9 ×1015 m Substituting value L and T in eq(3) M1L-1T-2 = 105 M1 = L1T2×105 M1 = 9×1015×(3×107)2 ×105=81×1035 kg Answer : (a)
Q.23
A string stretched with tension T has frequency 'f' which depends on length l and mass per unit length m then f ∝ ... .... [ Guj T.B.]
0%
a)
0%
b)
0%
c)
0%
d)
Explanation
Let f = k Talbmc Here let k be dimension less constant thus Dimension of LHS = Dimensions of RHS M0L0T-1 = [M1L1 T-2]a [L1]b[M1L-1]c On simplification a = 1/2, b = -1, c = -1/2 Answer : (b)
Q.24
A stationary object is accelerated. the distance d travelled by it in time t depends upon the acceleration a and time t. From above description, obtain the relation between d,a and t with the help of dimensional analysis , k= 1/2 .... ... [ Guj T.B. ]
0%
a) d = at
0%
b) d = (1/2)at2
0%
c) d = a/2t
0%
d) d = a2t2
Explanation
let d =k ap tq dimension of LHS = RHS L1 = [L1T-2]p [M0L0T1]q On simplification we get p= 1 and q = 2 also given k =1/2 thus option b is correct Answer : (b)
Q.25
The mass M of stone is flowing river depends on the speed v of a water, its density ρ and gravitational acceleration g. Then with the help of dimensional analysis, the mas of stone is ... times proportional to the power of the speed of water.... .... [ Guj T.B.]
0%
a) 2
0%
b) 3
0%
c) 6
0%
d) 8
Explanation
m = va ρb gc M1 = [LaT-a] [MbL-3b][LcT-2c] Comparing indices on both side we get 1 = b 0 = a-3b+c 0 = -a-2c On simplifying above equations we get a = 6, b = 1, c =-3 correct answer is (c) Answer : (c)
Q.26
The observation in experiment, length of rod are 2.51m, 2.56m, 2.49m then percentage error in length is ... ... [ Guj T.B.]
0%
a) 0.1%
0%
b) 1%
0%
c) 10%
0%
d) 0.01%
Explanation
Average reading = (2.51+2.56+2.49)/3 = 2.52 m Absoulte error of each observation Δl1 = 2.52 -2.51 = 0.01 Δl2 = 2.52 -2.56 = -0.04 Δl2 = 2.52 -2.49 = 0.03 Average absolute error, considering only magnitude Average absolute error = (0.01 + 0.04 + 0.03)/3 = 0.03 m Length of rod = (2.52 ±0.03) m Relative error = Absolute error/ average value = 0.03/2.52 = 0.01 Percentage error = Relative error ×100 Percentage error = 0.01 ×100 = 1% Answer : (b)
Q.27
If F = a√x + bt2 formula is dimensionally valid then what is the dimensional formula for 'a', here F= force and x = distance, t = time) ... .... [ Guj T.B.]
0%
a) M1L1T-1
0%
b) M1L1/2 T-2
0%
c) M1L1T-2
0%
d) M0L-1/2T2
Explanation
Use dimensional analysis Answer : (b)
Q.28
The periodic time T of the satellite revolving around the planet on circular orbit depends on a mass m, radius r and universal gravitational constant G, then T2 ∝ ... .. [ Guj T.B.]
0%
a) r1/2
0%
b) r3/2
0%
c) r
0%
d) r3
Explanation
T =k ma rb Gc Let k be dimensionless constant Dimension of LHS = RHS T1 = [Ma][Lb][M-cL3cT-2c] Comparing indices of MLT we get a = c = -1/2, b = 3/2 Thus T ∝ l3/2 T2 ∝ r3 Answer : (d)
Q.29
The formula for a physical quantity is here x, y,z are length . this physical quantity is ... ... [ Guj T.B. ]
0%
a) velocity
0%
b) length
0%
c) area
0%
d) volume
Explanation
Answer : (d)
Q.30
If E, M, L and G are energy, mass, angular momentum nad gravitational constnat respectively the dimensional formula of following equation will be ... ... [ Guj T.B]
0%
a) time
0%
b) angle
0%
c) mass
0%
d) length
Explanation
Dimensional formula for Energy = [M1 Angular momentum L = mvr = [M1L2T-1] gravitational constant G = [ M-1L3T-2 ] Mass =M = M1 Substitute dimensional formula in equation and solve, we get dimensions zero Now only angle do not have any dimension from given option thus angle is correct Answer : (b)
0 h : 0 m : 1 s
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
26
27
28
29
30
Report Question
×
What's an issue?
Question is wrong
Answer is wrong
Other Reason
Want to elaborate a bit more? (optional)