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BITSAT Mock Test - 5 - JEE MCQ

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30 Questions MCQ Test - BITSAT Mock Test - 5

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BITSAT Mock Test - 5 - Question 1

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A body of mass 0.5 kg travels in a straight line with velocity v = ax^3/2, where a = 5 m^-1/2 s^-1. The work done by the displacement from x = 0 to x = 2 m is?

A.
30 J
B.
40 J
C.
50 J
D.
60 J

Detailed Solution for BITSAT Mock Test - 5 - Question 1

When a variable force acts on a particle while it moves from point A to B, say along the path shown in the figure, work done by the force on the particle is given by

BITSAT Mock Test - 5 - Question 2

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A sonometer wire, 65 cm long, is in resonance with a tuning fork of frequency N. If the length of the wire is decreased by 1 cm and it is vibrated with the same tuning fork, then 8 beats are heard per second. What is the value of N?

A. 256 Hz
B. 384 Hz
C. 480 Hz
D. 512 Hz

Detailed Solution for BITSAT Mock Test - 5 - Question 2

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BITSAT Mock Test - 5 - Question 3

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A particle of charge -16 x 10^-18 C moving with velocity 10 ms^-1 along the x-axis enters a region where a magnetic field of induction B is along the y-axis and an electric field of magnitude 10⁴ Vm^-1 is along the negative z-axis. If the charged particle continues moving along the x-axis, then magnitude of B is

A.
10^-3 Wbm^-2
B.
10³ Wbm^-2
C.
10⁵ Wbm^-2
D.
10¹⁶ Wbm^-2

Detailed Solution for BITSAT Mock Test - 5 - Question 3

BITSAT Mock Test - 5 - Question 4

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A wave represented by the equation y = a cos (kx – ωt) is superposed with another wave to form a stationary wave such that the point x = 0 is a node. The equation of the other wave is

A.
y' = a sin (kx + ωt)
B.
y' = – a cos (kx – ωt)
C.
y' = – a cos (kx + ωt)
D.
y' = – a sin (kx – ωt)

Detailed Solution for BITSAT Mock Test - 5 - Question 4

To form a stationary wave, waves y and y' must travel in opposite directions.
Wave y = a cos (kx - ωt) travels along the positive x-direction.
Waves y' = -a cos (kx - ωt) and y' = -a sin (kx - ωt) in choices (2) and (4) travel along positive x-direction.
Hence choices (2) and (4) are not possible. Choice (1) is also incorrect becasue at x = 0
y' = a sin ωt and y = a cos (-ωt) = a cos ωt
Therefore, the resultant displacement at x = 0 which is y + y' = a sin ωt + a cos ωt is not zero, i.e. these waves do not produce a node at x = 0.
Choice (3) is correct because at x = 0, y + y' = 0.

BITSAT Mock Test - 5 - Question 5

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In a young double slit experiment, the intensity at a point where the path difference is (being the wavelength of the light used) is I. If I₀ denotes the maximum intensity, is equal to

A.
1/2
B.
3/4
C.
1/√2
D.
1/√3

Detailed Solution for BITSAT Mock Test - 5 - Question 5

As the path difference is
Corresponding phase difference is

if the Intensity of the source is
Then the intensity of the interference pattern is given by

BITSAT Mock Test - 5 - Question 6

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A tunnel is dug along the radius of Earth that ends at the centre. A body is released from the surface along the tunnel. The ball will bounce, after the first collision, at the centre up to a height of (radius of Earth is R and coefficient of restitution is e)

A. R
B. eR
C. e²R
D.

Detailed Solution for BITSAT Mock Test - 5 - Question 6

BITSAT Mock Test - 5 - Question 7

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The linear momentum p of a body of mass 2 kg varies with time t as p = 3t² + 4, where p and t are in SI units. It follows that the body is moving with a

A. constant speed
B. constant acceleration
C. variable acceleration
D. variable retardation

Detailed Solution for BITSAT Mock Test - 5 - Question 7

BITSAT Mock Test - 5 - Question 8

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Directions: The given figure shows the position-time (x-t) graph of one-dimensional motion of a body of mass 0.4 kg.

What is the magnitude of each impulse?

A.
0.2 Ns
B.
0.4 Ns
C.
0.8 Ns
D.
1.6 Ns

Detailed Solution for BITSAT Mock Test - 5 - Question 8

Between t = 0 and t = 2s, the speed of the body is v = slope of the (x - t) graph between t = 0 and t = 2s, i.e.,

At t = 2s, the velocity of the body is reversed and it moves in the opposite direction with a speed = -1 ms^-1. Therefore,
Impulse = change in momentum
= mv - (-mv) = 2mv
= 2 × 0.4 kg × 1 ms^-1
= 0.8 kg ms^-1 = 0.8 Ns
Hence, the correct choice is (3).

BITSAT Mock Test - 5 - Question 9

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A solenoid of inductance L and resistance R is connected to a battery. The time taken for the magnetic energy to reach of its maximum value is

Detailed Solution for BITSAT Mock Test - 5 - Question 9

The growth of current in an LR circuit is given by

where I₀ is the maximum current. The energy stored at time t is

We are required to find the time at which the energy stored is one-fourth the maximum value, i.e. when

i.e.
Using this in Eq. (1), we get

or
or log_e(2), which is choice (2).

BITSAT Mock Test - 5 - Question 10

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A block of weight 200 N is pulled along a rough horizontal surface at a constant speed by a force of 100 N, acting at an angle of 30° above the horizontal. The coefficient of friction between the block and the surface is

A.
0.43
B.
0.58
C.
0.75
D.
0.85

Detailed Solution for BITSAT Mock Test - 5 - Question 10

Since the block moves with a constant velocity, no net force acts on it. Therefore, the horizontal component F cos θ of force F must balance with the friction force, i.e. f_r = F cos θ. Also

Which is choice (2).

BITSAT Mock Test - 5 - Question 11

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The numbers of oxygen atoms shared per tetrahedron in
I. two dimensional sheet structured silicates
II. cyclic silicates and
III. single strand chain silicates
are

A.
3, 3, 2
B.
3, 2, 2
C.
4, 3, 1
D.
4, 3, 2

Detailed Solution for BITSAT Mock Test - 5 - Question 11

In two dimensional sheet silicates, three oxygen atoms each of unit are shared.

Cyclic silicates are obtained by sharing two oxygen atoms each of tetrahedron.

Chain silicates are formed by sharing of two oxygen atoms each of unit.

BITSAT Mock Test - 5 - Question 12

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The compounds formed at anode in the electrolysis of an aqueous solution of potassium acetate are

A. C₂H₆ and CO₂
B. C₂H₄ and CO₂
C. CH₄ and H₂
D. CH₄ and CO₂

Detailed Solution for BITSAT Mock Test - 5 - Question 12

This is an example of Kolbe's electrolysis.
2CH₃COOK 2CH₃COO^- + 2K⁺
At cathode:
2K⁺ + 2e^- 2K
2K + 2H₂O 2KOH + H₂
At anode:

BITSAT Mock Test - 5 - Question 13

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Which of the following substrates can be converted into CH₄and CH₃-CH₃ in a single step using different reactions?

A.
CH₃I
B.
CH₃OH
C.
CH₃CH₂I
D.
C₂H₅OH

Detailed Solution for BITSAT Mock Test - 5 - Question 13

BITSAT Mock Test - 5 - Question 14

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What is the shortest wavelength line in the Lyman series of the hydrogen spectrum? (R = 1.097 x 10^–2 nm^–1)

A.
94.21 nm
B.
91.16 nm
C.
911.6 nm
D.
93.6 nm

Detailed Solution for BITSAT Mock Test - 5 - Question 14

For Lyman series,
Hence, for to be the smallest, n should be the greatest (i.e. ).

= 1.097 x 10^-2 nm^-1

BITSAT Mock Test - 5 - Question 15

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Arrange the following in increasing order of their boiling points:
I. Ethylmethylamine
II. Propylamine
III. Trimethylamine

A.
II < I < III
B.
III < I < II
C.
III < II < I
D.
I < III < II

Detailed Solution for BITSAT Mock Test - 5 - Question 15

The correct order of boiling points is: Trimethylamine < Ethylmethylamine < Propylamine
This is because branched isomers have a lesser surface area and a lower boiling point.

BITSAT Mock Test - 5 - Question 16

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Directions: In the following problem, find from the four Response figures, the one which resembles the pattern formed when the Transparent sheet, carrying a design, is folded along the dotted line.

A.
A
B.
B
C.
C
D.
D

Detailed Solution for BITSAT Mock Test - 5 - Question 16

Hence, option 2 is correct.

BITSAT Mock Test - 5 - Question 17

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Directions: Four alternative summaries are given below the text. Choose the option that best captures the essence of the text.
The easiest way to do exercise is to make it a part of our daily life like walking or cycling instead of riding a car or bus and doing stretching exercises just for a while whenever we get a short break. I think nature is the best physical instructor. We can easily learn how to make exercise a part of ourselves just by watching the activities of birds, cats and dogs in our locality. I see that whenever a sparrow comes to sit on our windowsill, it gets itself busy with preening. One day, I had to shoo away a street dog that blocked our stairs but before moving away, it stretched its body to and fro and then walked away with dignity. A dog violently shakes its body to clean dirt and water. A cat does many kinds of acrobatic feats to lick its body clean from chest to tail!

A.
Daily exercise is as important as breathing and should be done as regularly.
B.
Exercise is a must for everyone; learn from nature.
C.
A strict routine is not required for work out.
D.
Animals have inculcated a routine of exercise in their normal daily tasks.

Detailed Solution for BITSAT Mock Test - 5 - Question 17

In the given passage, it is highlighted that exercise is crucial. It also highlights how various birds and animals take care of this aspect in their little activities - "I think nature is the best physical instructor." So, option (2) best explains it.

BITSAT Mock Test - 5 - Question 18

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Directions: In the following question, a series is given with one term missing. Choose the alternative that will complete the series.
119, 141, 165, 191, 219, ?

A.
254
B.
249
C.
248
D.
251

Detailed Solution for BITSAT Mock Test - 5 - Question 18

The series is following the pattern:
11² - 2 = 119
12² - 3 = 141
13² - 4 = 165
14² - 5 = 191
15² - 6 = 219
16² - 7 = 249

BITSAT Mock Test - 5 - Question 19

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Directions: Complete the series.
DEF, HIJ, MNO, ?

A.
STU
B.
RST
C.
RTV
D.
QRS

Detailed Solution for BITSAT Mock Test - 5 - Question 19

D + 4 = H, H + 5 = M, M + 6 = S
E + 4 = I, I + 5 = N, N + 6 = T
F + 4 = J, J + 5 = O, O + 6 = U

BITSAT Mock Test - 5 - Question 20

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Directions: In this question, the second figure of the Problem figures bears a certain relationship to the first figure. Similarly, one of the figures in the Answer figures bears the same relationship to the third figure. You have to select the figure from the set of answer figures which will replace the sign of question mark (?).

A.
(a)
B.
(b)
C.
(c)
D.
(d)

Detailed Solution for BITSAT Mock Test - 5 - Question 20

The figure rotates 45° in the anticlockwise direction. The blank circle attached to the head of the figure is changed from shaded to unshaded or vice-versa. Also, the angled line attached to the figure goes to the other side. After following this pattern, Answer figure (c) is obtained.

BITSAT Mock Test - 5 - Question 21

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Which colour does the teacher like?

A.
Blue
B.
Brown
C.
Purple
D.
Green

Detailed Solution for BITSAT Mock Test - 5 - Question 21

Rosy likes yellow and is a student. Mary and Andy like purple and blue, respectively, and neither of them is a teacher. The inspector likes blue, which means Andy is the inspector.

Lily likes brown. The librarian likes green.

Mary and Andy like purple and blue, respectively, and neither of them is a teacher. One of the five is a principal. Mary could not be the teacher, so Mary is the principal and Lily is the teacher.

So, the teacher likes brown colour.
Hence, option 2 is correct.

BITSAT Mock Test - 5 - Question 22

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Which of the following is a wrong pair?

A.
Andy - Inspector
B.
Purple - Principal
C.
Daisy - Green
D.
Rosy - Teacher

Detailed Solution for BITSAT Mock Test - 5 - Question 22

Rosy likes yellow and is a student. Mary and Andy like purple and blue, respectively, and neither of them is a teacher. The inspector likes blue, which means Andy is the inspector.

Lily likes brown. The librarian likes green.

Mary and Andy like purple and blue, respectively, and neither of them is a teacher. One of the five is a principal. Mary could not be the teacher, so Mary is the principal and Lily is the teacher.

'Rosy - Teacher' is a wrong pair. Rosy is a student.
Hence, option 4 is correct.

BITSAT Mock Test - 5 - Question 23

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Directions: Find the odd one out.

A. 120 - 560
B. 91 - 299
C. 78 - 169
D. 104 - 429

Detailed Solution for BITSAT Mock Test - 5 - Question 23

Except in the number pair 120 - 560, in all other number pairs, both the numbers are multiples of 13.
13 × 7 = 91, 13 × 23 = 299
13 × 6 = 78, 13 × 13 = 169
13 × 8 = 104, 13 × 33 = 429

BITSAT Mock Test - 5 - Question 24

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What value should come in place of the question mark (?) to complete the figure?

A.
15
B.
16
C.
17
D.
18

Detailed Solution for BITSAT Mock Test - 5 - Question 24

5 × 3 × 4 × 2 = = 12
5 × 6 × 3 × 2 = = 18
5 × 2 × 9 × 2 = = 18
Correct option is (4).

BITSAT Mock Test - 5 - Question 25

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Directions: The following question consists of a pair of words which have a certain relationship with each other. Select one pair from the four given alternatives, which has the same relationship as the original pair of words.
Wool : Warmth

A.
Spring : Elasticity
B.
Marketing : Manufacturing
C.
Person : Success
D.
Radio : Television

Detailed Solution for BITSAT Mock Test - 5 - Question 25

Second is the special property of the first. Wool is warm and spring is elastic.

BITSAT Mock Test - 5 - Question 26

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If y = 4x - 5 is a tangent to the curve y² = px³ + qat (2, 3), then

A. p = 2 and q = -7
B. p = -2 and q = 7
C. p = -2 and q = -7
D. p = 2 and q = 7

Detailed Solution for BITSAT Mock Test - 5 - Question 26

BITSAT Mock Test - 5 - Question 27

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If the equations x² − ax + b = 0 and x² + bx − a = 0 have a common root, then

A. a = b
B. a + b = -1
C. a + b = 1
D. a − b = 1

Detailed Solution for BITSAT Mock Test - 5 - Question 27

BITSAT Mock Test - 5 - Question 28

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The functionis

A. even
B. odd
C. periodic with period
D. Both (2) and (3)

Detailed Solution for BITSAT Mock Test - 5 - Question 28

BITSAT Mock Test - 5 - Question 29

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is equal to

A.
log_e(abcd)
B.
C.
D.
log (abcd)^1/4

Detailed Solution for BITSAT Mock Test - 5 - Question 29

=
⁼
=
= {log_e a + log_e b + log_e c + log_ed

BITSAT Mock Test - 5 - Question 30

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The roots of equation x⁶- 1 = 0 are

A.
-1 and 1
B.
1, ωand ω²
C.
-1, -ωand -ω²
D.
1, ω, ω², -1, -ωand -ω²

Detailed Solution for BITSAT Mock Test - 5 - Question 30

x⁶ - 1 = 0
(x³)² - 1 = 0
or, (x³ - 1)(x³ + 1) = 0
As we know, (x³ - 1) = 0
or, (x - 1)(x² + x + 1) = 0
∴ The roots are 1, ωandω².
Similarly when (x³ + 1) = 0
or, (x + 1)( x² - x + 1 ) = 0
Then x = -1, x = - ω and x = - ω²
Hence, x = 1, -1, ω, ω² , - ω and - ω²
(Where, ω = )

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