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COMEDK Mock Test - 7 - JEE MCQ


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30 Questions MCQ Test - COMEDK Mock Test - 7

COMEDK Mock Test - 7 for JEE 2024 is part of JEE preparation. The COMEDK Mock Test - 7 questions and answers have been prepared according to the JEE exam syllabus.The COMEDK Mock Test - 7 MCQs are made for JEE 2024 Exam. Find important definitions, questions, notes, meanings, examples, exercises, MCQs and online tests for COMEDK Mock Test - 7 below.
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COMEDK Mock Test - 7 - Question 1

A particle of mass m is fixed to one end of a light spring having force constant k and unstretched length l. The other end is fixed. The system is given an angular speed ω about the fixed end of the spring such that it rotates in a circle in gravity free space. Then the stretch in the spring is

Detailed Solution for COMEDK Mock Test - 7 - Question 1


At elongated position (x),
Fradial = mr ω2
∴ kx = m(ℓ + x)ω2

COMEDK Mock Test - 7 - Question 2

The circuit shown in the figure contains two diodes D1 and D2, each with a forward resistance of 50 ohms and infinite backward resistance. If the battery voltage is 6 V, the current (in amperes) through the 100 ohm resistance is

Detailed Solution for COMEDK Mock Test - 7 - Question 2

In the given circuit only diode D1 will allow the current to pass through as it is forward-biased.
Hence, the current through the 100 ohm resistance is (∵ resistance of D1 = 50 Ω)

= 0.02 A

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COMEDK Mock Test - 7 - Question 3

An elevator in a building can carry a maximum of 10 persons, with the average mass of each person being 68 kg. The mass of the elevator itself is 920 kg and it moves with a constant speed of 3 m/s. The frictional force opposing the motion is 6000 N. If the elevator is moving up with its full capacity, the power delivered by the motor to the elevator (g = 10 m/s2) must be at least:

Detailed Solution for COMEDK Mock Test - 7 - Question 3

F = (10 m + M)g + f
Where, m = 68 kg, M = 920 kg, f = 6000 N
⇒ F = 22,000 N
⇒ P = FV = 22,000 × 3 = 66,000 W

COMEDK Mock Test - 7 - Question 4

The truth table given below is for (A and B are the inputs, Y is the output). What type of GATE it will form?

Detailed Solution for COMEDK Mock Test - 7 - Question 4

Truth table with A and B as inputs and Y as output. We have to find which gate corresponds to the given truth table.
A. NOR gate can be made by using an OR gate followed by NOT gate , i.e. Given truth table does not satisfy this logic.

B. AND gate is a logic gate with the output as a product of the input, i.e.  The given truth table does not satisfy this logic.

C. XOR gate is a function of two logical variables A and B which gives the value 1 if one of the variables is zero and other is 1. If both the variables are 0 or both are 1, then the function is zero, i.e.,The given truth table does not satisfy this logic.

D. NAND gate is inverse of AND gate and is given by the expression 

Truth table for NAND gate is

This satisfies the given truth table.

COMEDK Mock Test - 7 - Question 5

A second pendulum is mounted in a space shuttle. Its period of oscillations will decrease when rocket is:

Detailed Solution for COMEDK Mock Test - 7 - Question 5

 

  • Time Period, T = 2π √(l/g')where,
    l = Length of seconds pendulum 
    g’ = Apparent Gravity
  • For the period of oscillations of Seconds Pendulum to decrease, the Apparent gravity (g’) has to increase because:
  • Hence, Time Period of oscillations of Seconds Pendulum will decrease when the rocket is ascending up with uniform acceleration.
COMEDK Mock Test - 7 - Question 6

The difference of speed of light in the two media A and B (vA - vB) is 2.6 × 107 m/s. If the refractive index of medium B is 1.47, then the ratio of refractive index of medium B to medium A is: (Given: Speed of light in vacuum, c = 3 × 108 ms-1)

Detailed Solution for COMEDK Mock Test - 7 - Question 6

Speed of light in a medium,

c is the speed of light and μ is the refractive index.
⇒ 
= 2.04 × 108 = 20.4 × 107 m/s
∴ vA - vB = 2.6 × 107 m/s
∴ vA = (20.4 + 2.6) × 107 = 23 × 107 m/s
∴  

COMEDK Mock Test - 7 - Question 7

An ideal transformer has 500 and the 1000 turns in the primary and the secondary coil. If the DC voltage of 120 V is applied to the primary coil, then the emf produced at the secondary coil will be:

Detailed Solution for COMEDK Mock Test - 7 - Question 7

Given: DC voltage EP = 120 V (Primary coil)

  • The transformer works on the principle of mutual inductance.
  • To induce emf in the secondary coil of the transformer, the magnetic flux associated with the secondary coil must change with respect to time.
  • When the DC voltage is applied at the primary coil of the transformer, the magnetic flux associated with the coil will remain constant with respect to time. So the emf will not induce at the secondary coil.
  • Therefore, when the DC voltage is applied at the primary coil, the induced emf in the secondary coil will be zero.
COMEDK Mock Test - 7 - Question 8

The energy released by the fission of one uranium atom is 200 MeV. The number of fissions required per second to produce 3.2 W of power is

Detailed Solution for COMEDK Mock Test - 7 - Question 8

The energy released by the fission of one uranium atom, E = 200 MeV
E = 200 × 10× 1.6 × 10-19 J
The number of fissions required per second, n/t = P/E.

COMEDK Mock Test - 7 - Question 9

The activity of a radioactive sample falls from 700 s-1 to 500 s-1 in 30 minutes. Its half life is close to:

Detailed Solution for COMEDK Mock Test - 7 - Question 9


 = 61.8 minute
⇒ T1/2  ≈ 62 minutes

COMEDK Mock Test - 7 - Question 10

A cylindrical solid of length L and radius a is having varying resistivity given by ρ = ρ0x where ρ0 is a positive constant and x is measured from the left end of the solid. The cell shown in the figure is having emf V and negligible internal resistance. The electric field as a function of x is best described by:

Detailed Solution for COMEDK Mock Test - 7 - Question 10
Consider an elemental part of solid at a distance xx from left end of width dxdx .

Resistance of this elemental part is,

COMEDK Mock Test - 7 - Question 11

An object of mass 5 kg is thrown vertically upwards from the ground. The air resistance produces a constant retarding force of 10 N throughout the motion. The ratio of time of ascent to the time of descent will be equal to [Use g = 10 ms-2]

Detailed Solution for COMEDK Mock Test - 7 - Question 11

COMEDK Mock Test - 7 - Question 12

For a gas, the difference between the two specific heats at constant pressure and constant volume is 4150 J kg-1 K-1 and their ratio is 1.4. What is the specific heat of the gas at constant volume in units of J kg-1 k-1?

Detailed Solution for COMEDK Mock Test - 7 - Question 12

Given: Cp - Cv = 4150 Jkg-1K-1 and Cp/Cv = 1.4 Or Cp = 1.4 Cv.

Therefore,

1.4 Cv - Cv = 4150

Or CV = 4150/0.4 = 10375 J kg-1 K-1

COMEDK Mock Test - 7 - Question 13

The magnifying power of a telescope with tube length 60 cm is 5. What is the focal length of its eye-piece?

Detailed Solution for COMEDK Mock Test - 7 - Question 13


For telescope,
Tube length (L) = fo + fes and magnification (m) = fe/fo , where fo and fe are focal lengths of objective and eye-piece
∴ fo + fe = 60 and fe = 5fo
∴ fo = 50 cm
fe = 10 cm

COMEDK Mock Test - 7 - Question 14

A compound A has molecular formula C₂Cl₃OH. It reduce Fehling's solution and on oxidation gives a monocarboxylic acid B. A is obtained by the action of Cl₂ on ethyl alcohol. A is

COMEDK Mock Test - 7 - Question 15

The heats of dissociation (k.cals/mole) of methane and ethane are 360 and 620 respectively. From these, C - C bond energy (k.cals/mole) in ethane can be evaluated as

COMEDK Mock Test - 7 - Question 16

When the densities of the ore particles and gangue particles are different, then the ore is concentrated by

COMEDK Mock Test - 7 - Question 17

Which of the following reacts rapidly with oxygen in air at ordinary temperature

COMEDK Mock Test - 7 - Question 18

Propene is reacted with HBr in presence of peroxide, the product is

COMEDK Mock Test - 7 - Question 19

Which one of the following compounds is not a vitamin?

COMEDK Mock Test - 7 - Question 20

Scurvy is caused due to deficiency of

COMEDK Mock Test - 7 - Question 21

The mass of 112 cm3 of CH₄ gas at STP is

COMEDK Mock Test - 7 - Question 22

In a reversible reaction, the catalyst

COMEDK Mock Test - 7 - Question 23

The IUPAC name of CH₃-C≡C-CH(CH₃)₂ is

COMEDK Mock Test - 7 - Question 24

Copper sulphate dissolves in excess of KCN to give

COMEDK Mock Test - 7 - Question 25

Let two points be A(1, -1) and B(0, 2). If a point P(x', y') be such that the area of ΔPAB = 5 sq. units and it lies on the line 3x + y - 4λ = 0, then a value of λ is

Detailed Solution for COMEDK Mock Test - 7 - Question 25

Given: 3x' + y' = 4λ

⇒ |x'(-3) - y'(1) + 1(2)| = 10
⇒ |-4λ + 2| = 10
⇒ 2 - 4λ = +10 or 2 - 4λ = -10
⇒ λ = -2 or λ = 3

COMEDK Mock Test - 7 - Question 26

In a triangle ABC, if  then the projection of the vector  is equal to:

Detailed Solution for COMEDK Mock Test - 7 - Question 26


Projection of 

COMEDK Mock Test - 7 - Question 27

If c is a point at which Rolle's theorem holds for the function f(x) = loge  in the interval [3, 4], where α ∈ R, then f''(c) is equal to

Detailed Solution for COMEDK Mock Test - 7 - Question 27

For application of Rolle's theorem,
f(3) = f(4)

COMEDK Mock Test - 7 - Question 28


Where [t] denotes greatest integer ≤ t. If m is the number of points where f is not continuous and n is the number of points where f is not differentiable, then the ordered pair (m, n) is

Detailed Solution for COMEDK Mock Test - 7 - Question 28


f is discontinuous but not diff. at x = 0

COMEDK Mock Test - 7 - Question 29

The line y = x + 1 meets the ellipse  at two points P and Q. If r is the radius of the circle with PQ as diameter, then (3r)2 is equal to:

Detailed Solution for COMEDK Mock Test - 7 - Question 29

Ellipse x2 + 2y2 = 4
Line y = x + 1
Point of intersection
x2 + 2(x + 1)2 = 4
3x2 + 4x - 2 = 0

m is slope of given line

COMEDK Mock Test - 7 - Question 30

Let . If  , then T + n(S) is equal to:

Detailed Solution for COMEDK Mock Test - 7 - Question 30

tanθ (sinθ + 1) - sin2θ = 0
tanθ (sinθ + 1 - 2cos2θ) = 0
⇒ tanθ = 0 or 2sin2θ + sinθ - 1 = 0
⇒ (2sinθ + 1) (sinθ - 1) = 0
⇒ sinθ = -1/2 or 1
But sinθ = 1 not possible

n(S) = 5

= 4

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