0 of 15 questions completed
Questions:
 1
 2
 3
 4
 5
 6
 7
 8
 9
 10
 11
 12
 13
 14
 15
Information
Quiz Description :
Name : AIPMT Practice Test – 1
Questions: 15
Maximum Marks: 60
Negative Marks: 1/4th
Time Allowed : 20 Minutes
Important for : AIPMT, State PMT and other medical entrance examinations.
You have already completed the Test before. Hence you can not start it again.
Test is loading...
You must sign in or sign up to start the Test.
You have to finish following quiz, to start this Test:
Congratulations!!!" AIPMT Practice Test  1 "
0 of 15 questions answered correctly
Your time:
Time has elapsed
Your Final Score is : 0
You have attempted : 0
Number of Correct Questions : 0 and scored 0
Number of Incorrect Questions : 0 and Negative marks 0
Average score  
Your score 

Not categorized
You have attempted: 0
Number of Correct Questions: 0 and scored 0
Number of Incorrect Questions: 0 and Negative marks 0
It’s time to share this quiz with your friends on Facebook, Twitter, Google Plus, Whatsapp or LinkedIn…
Click on View Questions Button to check Correct and incorrect answers.
AIPMT Practice Test 2 will be Available on 25th October 2015
Pos.  Name  Entered on  Points  Result 

Table is loading  
No data available  
 1
 2
 3
 4
 5
 6
 7
 8
 9
 10
 11
 12
 13
 14
 15
 Answered
 Review
 Question 1 of 15
1. Question
4 pointsA man weighs 80 kg. He stands on a weighing scale in a lift which is moved upward with a uniform acceleration of 5 m/s2. What would be the reading on the scale? Given (g= 10 m/s2)
CorrectHints When the lift is accelerated upward with acceleration a, then reading on the scale
R = m (g + a) = 80 (10 + 5) N = 1200 N
IncorrectHints When the lift is accelerated upward with acceleration a, then reading on the scale
R = m (g + a) = 80 (10 + 5) N = 1200 N
UnattemptedHints When the lift is accelerated upward with acceleration a, then reading on the scale
R = m (g + a) = 80 (10 + 5) N = 1200 N
 Question 2 of 15
2. Question
4 pointsA particle of mass m oscillate with SHM (Simple Harmonic Motion) between points x1 and x2, the equilibrium position being O. Its potential energy is plotted. It will be given below in the graph
CorrectPotential energy of particle performing SHM varies parabolically in such a way that at mean position it become zero and maximum at extreme position.
IncorrectUnattempted  Question 3 of 15
3. Question
4 pointsFuse wire is a wire of
CorrectA fuse wire is generally prepared from tinlead alloy (63% tin + 37% lead). It should have high resistivity and low melting point.
IncorrectA fuse wire is generally prepared from tinlead alloy (63% tin + 37% lead). It should have high resistivity and low melting point.
UnattemptedA fuse wire is generally prepared from tinlead alloy (63% tin + 37% lead). It should have high resistivity and low melting point.
 Question 4 of 15
4. Question
4 pointsA charge q is located at the center of a cube. The electric flux through any face is
CorrectThe total flux through the cube φ total = q/ϵo
So The electric flux through any face
Φ face = q/6 ϵo = 4πq/6 ( 4πϵo)IncorrectThe total flux through the cube φ total = q/ϵo
So The electric flux through any face
Φ face = q/6 ϵo = 4πq/6 ( 4πϵo)UnattemptedThe total flux through the cube φ total = q/ϵo
So The electric flux through any face
Φ face = q/6 ϵo = 4πq/6 ( 4πϵo)  Question 5 of 15
5. Question
4 pointsOne mole of an ideal gas at an initial temperature of T K does 6R joules of work adiabatically. If the ratio of specific heats of this gas at constant pressure and at constant volume is 5/3, the final temperature of gas will be
 Question 6 of 15
6. Question
4 pointsThe dimension of universal gravitational constant are
CorrectIncorrectUnattempted  Question 7 of 15
7. Question
4 pointsIn a PN junction photo cell, the value of the photo electromotive force produced by monochromatic light is proportional to
CorrectIn photocell, photo electromotive force, is the force that stimulate the emission of an electric current when photo voltaic action creates a potential difference between two points and the electric current depends on the intensity of incident light.
IncorrectIn photocell, photo electromotive force, is the force that stimulate the emission of an electric current when photo voltaic action creates a potential difference between two points and the electric current depends on the intensity of incident light.
UnattemptedIn photocell, photo electromotive force, is the force that stimulate the emission of an electric current when photo voltaic action creates a potential difference between two points and the electric current depends on the intensity of incident light.
 Question 8 of 15
8. Question
4 pointsA particle of mass m1 is moving with a velocity v1 and another particle of mass m2 is moving with a velocity v2. Both of them have the same momentum but their different kinetic energy are E1 and E2 respectively. If m1 > m2 then
CorrectKinetic energy = (m p^{2)}/2
IncorrectKinetic energy = (m p^{2)}/2
UnattemptedKinetic energy = (m p^{2)}/2
 Question 9 of 15
9. Question
4 pointsA bar magnet of length ‘l’ and magnetic dipole moment ‘M’ is bent in the form of an arc as shown in figure. The new magnetic dipole moment will be
CorrectIncorrectM = m × l (where m is strength of each pole of bar magnet and l is length)
When the bar magnet is bent in the form of an arc as shown in figure.Then
l = (π/3) r
or r = 3l/π
New magnetic dipole moment
M’ = m × 2r sin 30^{0 }= m2 (3l/π) (1/2) = 3ml/π = 3M/π  Question 10 of 15
10. Question
4 pointsIf the ratio of diameter, length and Young’s modules of steel and copper wires shown in the figure are p, q and s respectively, then the corresponding ratio of increase in their lengths would be
CorrectIncorrectY = FL/A ΔL
= 4 FL/ πD^{2}ΔL
So ΔL = 4FL/ πD^{2}Y
ΔL_{S}/ ΔL_{C} = (F_{S}/F_{C})(L_{S} /L_{C})(D_{C}^{2} /D_{S}^{2})(Y_{C} /Y_{S} ) Where S Steel and C Copper
F_{S} = (5m + 2m)g = 7mg
F_{C} = 5mg
Suppose (L_{S} /L_{C}) = q (D_{S} /D_{C}) = p (Y_{S} /Y_{C }) = S
ΔL_{S}/ ΔL_{C} = (7mg/5mg) (q) (1 /p)^{2} (1 /S ) = 7q/5sp^{2}
UnattemptedY = FL/A ΔL
= 4 FL/ πD^{2}ΔL
So ΔL = 4FL/ πD^{2}Y
ΔL_{S}/ ΔL_{C} = (F_{S}/F_{C})(L_{S} /L_{C})(D_{C}^{2} /D_{S}^{2})(Y_{C} /Y_{S} ) Where S Steel and C Copper
F_{S} = (5m + 2m)g = 7mg
F_{C} = 5mg
Suppose (L_{S} /L_{C}) = q (D_{S} /D_{C}) = p (Y_{S} /Y_{C }) = S
ΔL_{S}/ ΔL_{C} = (7mg/5mg) (q) (1 /p)^{2} (1 /S ) = 7q/5sp^{2}
 Question 11 of 15
11. Question
4 pointsAn electromagnetic wave of frequency v = 3.0 MHz passes from vacuum into a dielectric medium with relative permitivity ϵ = 4.0 Then
CorrectIncorrectFrequency of electromagnetic waves does not change with change in medium but wavelength and velocity of wave changes with changes in medium
Velocity of electromagnetic waves in vacuum
UnattemptedFrequency of electromagnetic waves does not change with change in medium but wavelength and velocity of wave changes with changes in medium
Velocity of electromagnetic waves in vacuum
 Question 12 of 15
12. Question
4 pointsThe output of OR gate is I
CorrectA B Y 0 0
1
1
0 1
0
1
0 1
1
1
From truth table we can observe that if either of input is one then output is one. Also if both the inputs are one the output is one.
IncorrectA B Y 0 0
1
1
0 1
0
1
0 1
1
1
From truth table we can observe that if either of input is one then output is one. Also if both the inputs are one the output is one.
 Question 13 of 15
13. Question
4 pointsFor angle of projection of a projectile at angle (45^{0 } ̶ θ) and (45^{0 } + θ), the horizontal range described by the projectile are in the ratio of
CorrectIncorrectHorizontal range (R) = (u^{2} sin 2θ)/g
For angle of projections (45^{0 } ̶ θ), the horizontal range is
So R_{1} = [u^{2} sin {2 (45^{0 } ̶ θ )}]/g = [u^{2} sin {90^{0 } ̶ 2θ }]/g = u^{2} cos 2θ/g
For angle of projection (45^{0 } + θ), the horizontal range is
R_{2} = [u^{2} sin {2 (45^{0 } + θ )}]/g = [u^{2} sin {90^{0 } + 2θ }]/g = u^{2} cos 2θ/g
So R_{1} /R_{1} = (u^{2} cos 2θ/g )/ (u^{2} cos 2θ/g ) = 1/1
UnattemptedHorizontal range (R) = (u^{2} sin 2θ)/g
For angle of projections (45^{0 } ̶ θ), the horizontal range is
So R_{1} = [u^{2} sin {2 (45^{0 } ̶ θ )}]/g = [u^{2} sin {90^{0 } ̶ 2θ }]/g = u^{2} cos 2θ/g
For angle of projection (45^{0 } + θ), the horizontal range is
R_{2} = [u^{2} sin {2 (45^{0 } + θ )}]/g = [u^{2} sin {90^{0 } + 2θ }]/g = u^{2} cos 2θ/g
So R_{1} /R_{1} = (u^{2} cos 2θ/g )/ (u^{2} cos 2θ/g ) = 1/1
 Question 14 of 15
14. Question
4 pointsA microscope is focused on a mark on a piece of paper and then a slab of glass of thickness 3 cm and refractive index 1.5 is placed over the mark. How should the microscope be moved to get the mark in focus again
CorrectIncorrectApparent depth = real depth/μ = 3/1.5 = 2cm
UnattemptedApparent depth = real depth/μ = 3/1.5 = 2cm
 Question 15 of 15
15. Question
4 pointsA dog while barking delivers about 1 mW of power . If this power is uniformly distributed over a hemispherical area, what is the sound level at a distance of 5m? What would the sound level be if instead of 1 dog, 5 dogs start barking at the same time each delivering 1 mW of power
CorrectIncorrectAs power is distributed uniformly in a hemisphere, intensity at a distance of 5 m from the source will be
I = P/S = P/(1/2) 4πr^{2 }= 10^{ ̶ 3}/2π5^{2} = 6.37 μW/m^{2}
SL = 10 log (I/I_{0}) = 10 log (6.37 × 10 ^{̶ 6}/10 ^{̶ 12})
SL = 10 [ log 6.37 + 6 log 10] = 10 [0.80 + 6]
SL = 68 dB
If there are 5 dogs barking at the same time and same level, I_{2} = 5 I_{1 }. so
SL_{2 } ̶ SL_{1} = 10 log (I_{2} /I_{1}) = 10 log (5I_{1} /I_{1})
So SL_{2} = SL_{1} + 10 log 5 = 68 + 10 × 0.7 = 75 dB
UnattemptedAs power is distributed uniformly in a hemisphere, intensity at a distance of 5 m from the source will be
I = P/S = P/(1/2) 4πr^{2 }= 10^{ ̶ 3}/2π5^{2} = 6.37 μW/m^{2}
SL = 10 log (I/I_{0}) = 10 log (6.37 × 10 ^{̶ 6}/10 ^{̶ 12})
SL = 10 [ log 6.37 + 6 log 10] = 10 [0.80 + 6]
SL = 68 dB
If there are 5 dogs barking at the same time and same level, I_{2} = 5 I_{1 }. so
SL_{2 } ̶ SL_{1} = 10 log (I_{2} /I_{1}) = 10 log (5I_{1} /I_{1})
So SL_{2} = SL_{1} + 10 log 5 = 68 + 10 × 0.7 = 75 dB