bVQhrpracticequestions1_25_mcqs
-
Upload
shikari-shikarwala -
Category
Documents
-
view
216 -
download
0
Transcript of bVQhrpracticequestions1_25_mcqs
-
8/7/2019 bVQhrpracticequestions1_25_mcqs
1/17
Practice Questions and Solutions
Q1. Which of the following pairs of quantities have same dimensions?
A. acceleration and angular acceleration
B. stress and strain
C. latent heat and specific heat
D. momentum and impulse
E. mass and weight
Ans. (D) is correct choice. Momentum= mass x velocity =[MLT-1]
Impulse=Force x time = [MLT-2] x [T]
= [MLT-1
]
Choices A, B, C and E are incorrect.
Q2. Dimensional formula of Gravitational constant is
A. [M-1L2T]
B. [M-1L3T-2]
C. [M0L3T-2]
D. [M-1L-1T-2]
E. [M 1 L1 T 1]
-
8/7/2019 bVQhrpracticequestions1_25_mcqs
2/17
Ans. (B) is the correct choice,
Gravitational constant G=F x r2/m1 x m
Therefore dimension of G= [MLT-2][L2]/[M2]
= [M-1L3T-2]
Q3. Plancks constant has dimension of
A. Energy
B. Angular momentum
C. Power
D. momentum
E. Impulse.
Ans. (B) is the correct choice.
Plancks constant h=E/= Energy X Time
Dimension of h = [ML 2 T -2][T]
= [ML 2 T -1]
And angular momentum=Linear momentum X distance
Therefore dimension of angular momentum= [M L T -1][L]
= [M L 2 T 1]
Q4. In the fig. shown above two objects are approaching some point Owith equal velocities v. What is the magnitude relative velocity of
object 1 with respect to object 2?
-
8/7/2019 bVQhrpracticequestions1_25_mcqs
3/17
(A). 2 v
(B). 2 v
(C). v/2
(D). v/2
(E). v
Ans. (B) is the correct choice. Magnitude of relative velocity of object 1with respect to object 2 is v12= (v 2 + v 2)1/2 =2 v
Q5.The velocity of a particle is v= 5+(x + yt), where a and b are
constants and t is the time. The acceleration of the particle is
(A). x
(B). y
(C). 0
(D). xy
-
8/7/2019 bVQhrpracticequestions1_25_mcqs
4/17
(E) x/y
Ans. (B) is the correct choice.
v = 5 + x + yt
Acceleration dv/dt=d/dt (5 + x + yt) =y
Q6. The following forces are acting simultaneously on a particle
(1). (2 + 3 2k) N
(2). (3 + 3k) N
(3). (-5 -2 -3k) N
So the particle will move in
(A). X-Y Plane
(B). X Plane
(C). Y-Z Plane
(D). Y Plane
(E). X-Z Plane
Ans. (C). is the correct choice.
The total force acting on the particle is the sum of three forces i.e.
F= (2 +3 -2 k) + (3 + -3 k) + (-5 -2 +k)
F = 2-4 k
Therefore the particle will move in Y-Z Plane.
Options (A), (B), (D), (E) are incorrect, because F= 2 -4 k.
is a unit vector in Y direction and k is a unit vector in Z direction. So,particle will move in Y-Z Plane.
-
8/7/2019 bVQhrpracticequestions1_25_mcqs
5/17
Q7 A projectile is fired from point O at an angle with the
horizontal. What will be the angle between instantaneous velocity and
acceleration at the highest point h?
(A). 45 0
(B). 90 0
(C). 0 0
(D). 180 0
(E). 30 0
Ans. (B) is the correct choice. At point h, acceleration is directed in thedownward direction and velocity is along the tangent at that point. So, the
angle between the two is 90 0.
-
8/7/2019 bVQhrpracticequestions1_25_mcqs
6/17
Hence the Options (A), (C), (D), (E) are incorrect.
Q8. If y=a +bt +ct2, where y is in meter and t is in second. The unit of
b is
(A). m
(B). s
(C). ms -1
(D) ms -2
(E). m -1s -1
Ans. (C) is the correct choice. According to dimensional analysis
[y]= [bt]
[L]=b [T]
b= [L]/ [T] or unit of b=ms -1
-
8/7/2019 bVQhrpracticequestions1_25_mcqs
7/17
Q9. The above fig. shows displacement time graph of a body in a
straight like. We can conclude from the graph that
(A). the velocity increases uniformly
(B). the velocity is constant.
(C). the velocity decreases uniformly.
(D). the body is subjected to acceleration from O to A.
(E). the velocity of the body at A is zero
Ans. (E) is the correct choice.
Velocity at any point of the graph is found by drawing tangent to that point.At point A, tangent to the curve is parallel to time axis. It shows that velocity
at A is zero.
option (A),(B),( C),(D) are incorrect since the displacement is first
increasing with time and then starts decreasing, so the velocity of the bodycan not increase or decrease uniformly and nor can it be constant.
-
8/7/2019 bVQhrpracticequestions1_25_mcqs
8/17
Q10. A ball is dropped from the top of a tower of height h; meters. It
takes T seconds to reach the ground. Where is the ball at the time T/2
second?
(A). at h/2 meter from the ground.
(B). at h/4 meters from the ground.
(C). at 3h/4meter from the ground.
(D).at h/8 meter from the ground.
(E). at h/3 meter from the ground.
Ans. (C) is the correct choice.
h=ut+1/2gT2
The ball is dropped from rest, so u=0
So, h=1/2gT 2
After T/2 seconds, h1=1/2g (T/2) 2
h1=1/2g T2 /4
h1=1/4(1/2gT 2)
h1= h/4
So after T/2 sec. height above the ground =h -h/4=3h/4
Q11.Two resistances when connected in series become 6 ohms and 1
ohm when connected in parallel. The two resistances are
(A). 2, 4(B). 1, 5
(C). 1, 4
(D). 4.7 ,1.3
-
8/7/2019 bVQhrpracticequestions1_25_mcqs
9/17
(E). 2.5, 3.5
Ans. (D) is the correct choice.
R1+R2=6
R1 R2/R1 +R2=1
Therefore R1 R2/6=1
R1 R2=6
(R1 - R2)2 = ( R1 + R2 )
2 - 4R1 R2=36-24
R1 - R2=2 3
Therefore R1= 4.7R2= 1.3
Q12. Three voltmeters A, B, C are having resistances 2R, 3R, 6R
respectively, are connected as shown. When some potential difference is
applied between X&Y, the voltmeter readings are V A, V B and V Crespectively. Then,
-
8/7/2019 bVQhrpracticequestions1_25_mcqs
10/17
(A). V A = V B= V C
(B). V A V B = V C
(C).V A =V B V C
(D). V A V B V C
(E). V A V B =V C
Ans. (A) Is the correct choice.
From ohms law V=IR
V A=I.2R=2IR
The division of current in the two parallel branches will be according to the
resistance of each branch.
Current I B = 2/3 I
And I C =1/3 I
Therefore V B= I B RB =2/3 I x 3R=2 IR
AND V C = I C RC =1/3 I x 6R=2IR
Therefore V A= V B= V C
Q13. If the light is incident on a glass slab at polarizing angle then
A). Reflected light is completely polarized.
B). Refracted light is completely polarized.
C). both are completely polarized.
D). both are partially polarized.
-
8/7/2019 bVQhrpracticequestions1_25_mcqs
11/17
E) Both are un polarized.
Ans. (A) is the correct choice.
When light is incident at polarizing angle, it is partially reflected and
partially refracted. The reflected light is completely polarized.
Q14. The minimum resistance that can be obtained with ten 0.1
resistors is
(A). 1
(B). 0.01
(C). 0.1
(D). 0.001
(E).10
Ans. (B) is the correct choice.
Ten resistors connected in parallel with given, minimum resistance=0.01
Q15. A battery of e.m.f E and internal resistance r is being charged with
current I. The terminal potential will be
(A). E +I r
B). E I r
C).E
D).I r
E).none of the above
Ans. (A) is the correct choice
When the cell is charged, the terminal potential is more than the e.m.fand is equal to E +I r
-
8/7/2019 bVQhrpracticequestions1_25_mcqs
12/17
Option (B) is incorrect because E I r is the potential when the cell isbeing discharged i.e. when the current is being drawn from the cell.
Options (C) and (E) are also incorrect.
Q16. If potential difference between two plates separated by a
distance d is V, then the electric field is equal to
A).1/Vd
B). V d
C). V/d
D). d/V
E). V 2 d
Ans. (C) is the correct choice.
The electric field E between two plates at a potential difference Vseparated by a distance d is V= E/d
Q17. In the above fig. 9, what is the e.m.f of the given battery if
the e.m.f of each cell is 2 V.?
-
8/7/2019 bVQhrpracticequestions1_25_mcqs
13/17
A).12V
B). 10V
C). 8V
D). 6V
E). 4V
Ans. (E) is the correct choice.
The e.m.f of four cells is in the same direction and e.m.f of two cells is inopposite direction.
Therefore total e.m.f = (2 x4) 2 x 2=8 -2 =4V.
Q18. Intensity of light depends on its
A). Velocity
B).Wavelength
C).Amplitude
D).Frequency
E).Density
-
8/7/2019 bVQhrpracticequestions1_25_mcqs
14/17
Ans. (C) is the correct choice.
According to wave theory intensity of light I A 2 (square of amplitude)
Q19. Properties which affect the speed of light in space are
A). only mechanical
B). only electrical
C). only magnetic
D). electrical and magnetic
E). electrical and mechanical
Ans. (D) is the correct choice.
Light waves are electromagnetic. They are produced due to variation ofelectric and magnetic fields. So the properties which affect their speed are
electrical and magnetic.
Options (A), (B), (C),(E) are incorrect.
Q20. If the source of light in Youngs double slit experiment is changed
from red to green then fringes will become
A). Darker
B). Brighter
C). Thinner
D). Broader
E). Remain same
Ans. (C) is the correct choice.
-
8/7/2019 bVQhrpracticequestions1_25_mcqs
15/17
Fringe width, = D/d. With change in color there is a change inwavelength of light.
Fringe width Wavelength.
So as the wavelength decreases the fringe width also decreases. Since thefringe width decreases the fringes become thinner.
Options (A), (B), (D), (E) are incorrect.
Q21.For obtaining diffraction pattern the aperture of the slit should be
of the order of
A). B) ./2
C). /4
D). 2
E). 2
Ans. (A) is the correct choice.
Condition for diffraction to take place is that the aperture of the slit should
be of the order of the wavelength of light.
Q22. Ordinary light is
A). Plane polarized
B). circularly polarized
C). partially polarized
D). unpolarized
E). elliptically polarized.
Ans. (D) is the correct choice.
-
8/7/2019 bVQhrpracticequestions1_25_mcqs
16/17
Ordinary light is unpolarized. E.g. sunlight or light coming out of a bulb.
Q23. In the interference experiment spacing between successive maxima
or minima is
A). Dd/
B). D/d
C).d D
D). /dD
E). d/D
Ans. (B) is the correct choice.
Fringe width due to interference is = D/d
Q24. Two waves have intensities in the ratio 1:9. After interference ratio
of maximum to minimum intensity will be
A). 1:3
B). 3:1
C). 4:1
D). 1:4
E). 9:1
Ans. (C) is the correct choice.
Intensity (amplitude) 2
Given I 1 /I 2 =1/9
A 12/ A 2
2 =1/9
A 1 / A 2 =1/3
-
8/7/2019 bVQhrpracticequestions1_25_mcqs
17/17
Now, I max. / I min. = (a 1 + a 2 )2 / (a 1- a 2)
2
= (3+1) 2/ (3-1) 2
=16/4
So, I max. / I min = 4:1