MCQGeeks
0 : 0 : 1
CBSE
JEE
NTSE
NEET
English
UK Quiz
Quiz
Driving Test
Practice
Diagrams
Games
NEET
Physics NEET MCQ
Quiz 4
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
The poisson ratio can not have the value [ EAMCET 1989]
0%
a) 0.7
0%
b) 0.2
0%
c)0.1
0%
d)0.5
Explanation
Answer:(a)
Q.2
Streamline flow is more likely for liquids with
0%
a) High density an low viscosity
0%
b) Low density and high viscosity
0%
c)High density and high viscosity
0%
d)Low density and low viscosity
Explanation
Answer: (b)
Q.3
Two cylinder A and B are made of the same material. the length and radii of the two cylinder are in the ratio of 1 :Both are twisted by the same external torque. the ratio of the angle of twist of A and B is
0%
a) 1: 2
0%
b) 1 : 4
0%
c)1 : 8
0%
d)8 : 1
Explanation
torque required to prode twist of Φ in a wire of length 'l', radius 'r' and modulus of rigidity η is given by equationGiven torque is same , material is same r2 / r1=2 l2 / l1=2 from above formula and taking ratios we get Answer:(d)
Q.4
Two cylinder A and B of the same material have same length, the radii of A and B are in the ratio of 1:The two are joined end to end as shown in figure. The upper end of A is rigidly fixed. The lower end of B is twisted through an angle θ, the angle of twist of cylinder A is
0%
a) θ
0%
b) (15/16) θ
0%
c) (16/17)θ
0%
d) (17/16) θ
Explanation
The twist at upper end of A=0. the twist at the point is Φ, the twist at the lowest end=θ. Then,given τA=τB Answer: (c)
Q.5
In steel the Young's modulus and strain at the breaking point are 2×1011Nm-2 and 0.15 respectively. The stress at the break point for steel is therefore [ MPPET 1990]
0%
a)1.33 × 1011Nm-2
0%
b) 1.33×1012Nm-2
0%
c)7.5×10⁻³ Nm-2
0%
d)3×1010Nm-2
Explanation
We know Y=stress / strain Stress=Y ( strain) Stress=(0.15) (2×1011)=3×1010Nm-2 Answer: (d)
Q.6
Which of the following statement is correct? [ MPPET 1992]
0%
a) Hook's law is applicable only within elastic limit
0%
b) The adiabatic and isothermal elastic constant of gas are equal
0%
c)Young's modulus is dimensionless
0%
d)Stress multiplied by strain is equal to the stored energy
Explanation
Answer: (a)
Q.7
The force required to stretch a steel wire of 1cm2 cross-section to 1.1 times its length would be ( Y=2×1011 Nm-2) MPPER 1992]
0%
a) 2×106 N
0%
b) 2×103 N
0%
c)2×10⁻⁶
0%
d)2×10⁻⁷ N
Explanation
Increase in length=1.1(l) - l=0.1lFrom the formula of Young's modulus Answer:(a)
Q.8
Which of the following substance possesses the highest elasticity [ MPPMT 1992]
0%
a) Rubber
0%
b) Glass
0%
c) Steel
0%
d) Copper
Explanation
Answer: (c)
Q.9
Which of the following quantities does not have the unit of force per unit area [ MPPMT]
0%
a)Stress
0%
b) Strain
0%
c)Young's modulus of elasticity
0%
d)Pressure
Explanation
Answer: (b)
Q.10
A rod of length l and radius is joined to a rod of length l/2 and radius r/2 of same material. The free end of small rod is fixed to rigid base and the free end of larger rod is given a twist of θ°, the twist angle at the joint will be
0%
a) θ/4
0%
b) θ/2
0%
c)(5/6) θ
0%
d)(8/9) θ
Explanation
Twist at upper end of small rod A=0Twist at the joint is=ΦTwist at lower end of big rod B=θ since torque acting on the rods is same Answer: (d)
Q.11
As steel wire has length 2.0 m, radius 1.0 mm and Y=20×1010 N/mA sphere of mass 10 kg is attached to one end of the wire ,which is then whirled in a vertical circle with an angular velocity of 2 revolutions per sec. The elongation of wire when the mass is at lowest point of the path is nearly [ AMU 1995]
0%
a)1.0 mm
0%
b) 2.0 mm
0%
c)0.1 mm
0%
d)0.01 mm
Explanation
Tension at lower end=Centrifugal force + weight of sphereT=mlω2 + mg here m=10 kgω=2 revolution/sec.l=2.0 m T=10×2×(2π×2) + 10×10 T=420 N Answer:(a)
Q.12
Two wires of the same material have lengths in ratio 1:2 and their radii are in the ratio 1 :√If they are stretched by applying equal force, the increase in their lengths will be in the ratio.. [ MPPMT 1994]
0%
a) 2 : √2
0%
b) √2 : 2
0%
c) 1 : 1
0%
d) 1 : 2
Explanation
We know that increase in length is given by formula Given that material same so Y is same, Force is same for both wire Answer: (c)
Q.13
A uniform cube is subjected to volume compression. If each side is increased by 1% then bulk strain is .. [ EAMCET 1995]
0%
a)0.01
0%
b) 0.06
0%
c)0.02
0%
d)0.03
Explanation
Strain=ΔV/V=(3Δl)/ lGiven increase in length h=1%=0.01 Bulk Strain=3 (0.01)=0.03Answer: (d)
Q.14
A cylinder tree has a breaking stress of 106 N/mThe density of the tree is 2×104 kg/mThe maximum possible height of the tree is then [ CET 1994]
0%
a) 25 m
0%
b) 150 m
0%
c)5.0 m
0%
d)100 m
Explanation
Breaking stress=hdgd=density , h is height 106=h ×2×104×10 h=5 m Answer: (c)
Q.15
A force of 103 N stretches the length of a hanging wire by 1 millimeter. The force required to stretch a wire of same material and length but having four times the diameter by 1 millimetre is [ PMT 1995]
0%
a) 4×103 N
0%
b) 16×103 N
0%
c)(1/4)×103 N
0%
d)(1/16)×103 N
Explanation
Increase in length is given by Same material hence Y is constant, Length same, but diameter is four times Answer:(b)
Q.16
When a weight of 10 kg is suspended from a copper wire of length 3 metres and diameter 0.4 mm, it length increases by 2.4 cm. If the diameter of the wire is doubled, then the extension in its length will be [ MPPMT 1994]
0%
a) 9.6 cm
0%
b) 4.8 cm
0%
c) 1.2 cm
0%
d) 0.6cm
Explanation
Increase in length is given by force is same in both the cases, length is same , material is same only diameter is doubled Answer: (d)
Q.17
A 2 m long rod of radius 1cm, which is fixed from one end is given twist of 0.8 radians. The shear strain developed will be [ Raj.PET 1997]
0%
a) 0.002
0%
b) 0.004
0%
c)0.008
0%
d)0.016
Explanation
we know that angle of shear θ=Φr / l Here Φ is angle of twist θ=(0.8×0.01) / 2=0.004 radianAnswer: (b)
Q.18
Two wires of same material and length but diameter in the ratio 1 : 2 are stretched by the same force. the potential energy per unit volume for the two wires when stretched will be in the ratio [ CPMT 1998]
0%
a) 16 : 1
0%
b) 4 : 1
0%
c) 2 : 1
0%
d) 1 : 1
Explanation
Potential energy per unit volume From the formula for Young's modulus Substituting value of Δl in equation for Potential energy we get same material hence Y is same, Length is also same , force is also same, but diameter is doubled there fore radius is doubled thus Answer: (a)
Q.19
A fixed volume of iron is drawn in to a wire of length l. The extension x produced in the wire by a constant force F is proportional to [ MPPMT 1999]
0%
b) 1/l
0%
c)l2
0%
d)l
0%
a)1 / l²
Explanation
Force is constant and volume is fixeed then Extension ∝ (length)2 Answer: (c)
Q.20
The elastic energy stored in a wire of Youngs modulus Y is [ MPPMT 1999]
0%
a) Y×(strain)2 / volume
0%
b) stress × strain × volume
0%
c)(stress)2 × volume / 2Y
0%
d)½ Y × stress × strain × volume
Explanation
Enegy E=½ × stress × strain × volume Strain=Stress / Y ∴ E=½ × Stress × (stress/Y) × volume E=(stress)2 × volume / 2YAnswer: (c)
Q.21
When there are no external force, the shape of liquid drop is determined by [ CPMT 1999]
0%
a) surface tension of the liquid
0%
b) Density of liquid
0%
c)viscosity of liquid
0%
d)temperature of air only
Explanation
Answer:(a)
Q.22
An iron needle slowly placed on the surface of water floats on it because [ MNR 1993]
0%
a) When inside the water it will displace water more than its weight.
0%
b) The density of the material of needle is less than that of water
0%
c) its surface tension
0%
d) Of its shape
Explanation
Answer: (c)
Q.23
A number of water droplets each of radius r collusion to form a droplet of radius R, the rise in temperature dθ is : [ MPPMT 1994]
0%
a)
0%
b)
0%
c)
0%
d)
Explanation
Let n be the number of drop coalesce to form a droplet of radius R, Work done W=T×ΔA W=T×4π[nr2 - R2] we know that R3=n r3 Thus n=R3 / r3 We know that W=JH And H=m×s×Δθ Here s is specific heat , Δθ is increase in temperature , m is mass mass m=(4/3) πR3 ×density=(4/3) πR3 ×1 W=J (4/3) πR3×1×1×Δθ Thus comparing formulas for Work we get Answer: (c)
Q.24
More liquid rises in a thin tube because of [ CPMT 1987]
0%
a) Larger value of radius
0%
b) Larger value of surface tension
0%
c) small values of surface tension
0%
d) small value of radius
Explanation
h ∝ 1/r Answer: (d)
Q.25
Two spherical soap bubbles of radii r1 and r2 in vacuum collapses under isothermal conditions. the resulting bubble has a radius R such that
0%
a)
0%
b)
0%
c)
0%
d)
Explanation
When two soap bubbles coalesce under isothermal condition to form a new bubble, then ∑PV=constant Then P1V1 + P2V2=PV Pressure inside the bubble=4T/r Answer: (c)
Q.26
When soap bubble is charged [ Raj.PMT 1997]
0%
a) It contracts
0%
b) It expands
0%
c)It does not undergo any change in size
0%
d)None of these
Explanation
Answer: (b)
Q.27
Small droplets of a liquid are usually more spherical in shape than the larger drop of the same liquid because [ EAMCET 1988]
0%
a)Force of surface tension is equal and opposite to the force of gravity
0%
b)Force of surface tension predominates the force of gravity
0%
c)Force of gravity predominates the force of surface tension
0%
d)Force of gravity and force of surface tension act in the same direction and are equal
Explanation
Answer:(b)
Q.28
With a rise in temperature, the surface tension of liquid [ MPPMT 199]
0%
a) Increases
0%
b) Decreases
0%
c) Does not change
0%
d) Changes erratically
Explanation
Answer: (b)
Q.29
Water rises to a height of 10 cm in a capillary tube, and mercury falls to a depth of 3.42 cm in the same capillary tube. if the density of mercury is 13.6 and the angle of contact is 135°, the ratio of surface tension for water and mercury is [ MPPMT 1988]
0%
a)1 : 0.5
0%
b) 1 : 3
0%
c)1 : 6.5
0%
d)1.5 : 1
Explanation
formula for height in capillary tube Here d=density of liquid, θ=angle of contact, r=radius of tubeOn taking the ratioAnswer: (c)
Q.30
Surface tension may be defined as
0%
a) The work done per unit area in increasing the surface area of liquid under isothermal condition
0%
b) The work done per unit area in increasing the surface area of a liquid under adiabatic condition
0%
c) the work done per unit area in increasing the surface area of liquid under both isothermal and adiabatic conditions
0%
d)free surface energy per unit volume
Explanation
Answer: (c)
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)