JEE Exam  >  JEE Tests  >  WBJEE Sample Papers, Section Wise & Full Mock Tests 2025  >  WBJEE Previous Year (2010) - JEE MCQ

WBJEE Previous Year (2010) - JEE MCQ


Test Description

30 Questions MCQ Test WBJEE Sample Papers, Section Wise & Full Mock Tests 2025 - WBJEE Previous Year (2010)

WBJEE Previous Year (2010) for JEE 2024 is part of WBJEE Sample Papers, Section Wise & Full Mock Tests 2025 preparation. The WBJEE Previous Year (2010) questions and answers have been prepared according to the JEE exam syllabus.The WBJEE Previous Year (2010) MCQs are made for JEE 2024 Exam. Find important definitions, questions, notes, meanings, examples, exercises, MCQs and online tests for WBJEE Previous Year (2010) below.
Solutions of WBJEE Previous Year (2010) questions in English are available as part of our WBJEE Sample Papers, Section Wise & Full Mock Tests 2025 for JEE & WBJEE Previous Year (2010) solutions in Hindi for WBJEE Sample Papers, Section Wise & Full Mock Tests 2025 course. Download more important topics, notes, lectures and mock test series for JEE Exam by signing up for free. Attempt WBJEE Previous Year (2010) | 70 questions in 140 minutes | Mock test for JEE preparation | Free important questions MCQ to study WBJEE Sample Papers, Section Wise & Full Mock Tests 2025 for JEE Exam | Download free PDF with solutions
WBJEE Previous Year (2010) - Question 1

Experimental investigations show that the intensity of solar radiation is maximum for a wavelength 480 nm in the visible region. Estimate the surface temperature of sun. Given Wein’s constant b = 2.88 × 10–3 mK.

Detailed Solution for WBJEE Previous Year (2010) - Question 1

λm × T = b
λm = 480 nm

WBJEE Previous Year (2010) - Question 2

The temperature of an ideal gas is increased from 120 K to 480 K. If at 120 K, the root mean square speed of gas molecules is v, then at 480 K it will be

Detailed Solution for WBJEE Previous Year (2010) - Question 2


1 Crore+ students have signed up on EduRev. Have you? Download the App
WBJEE Previous Year (2010) - Question 3

Two mirrors at an angle θº produce 5 images of a point. The number of images produced when θ is decreased to θº – 30º is

Detailed Solution for WBJEE Previous Year (2010) - Question 3

No. of images = 5
∴ θ = 60º
New angle = θ – 30º = 30º. No of images 

WBJEE Previous Year (2010) - Question 4

he radius of the light circle observed by a fish at a depth of 12 meter is (refractive index of water = 4/3)

Detailed Solution for WBJEE Previous Year (2010) - Question 4

WBJEE Previous Year (2010) - Question 5

In Young’s double slit experiment, the fringe width is β. If the entire arrangement is placed in a liquid of refractive index n, the fringe width becomes :

WBJEE Previous Year (2010) - Question 6

A plano-convex lens (f = 20 cm) is silvered at plane surface. Now focal length will be :

Detailed Solution for WBJEE Previous Year (2010) - Question 6


P = 2PL + PM
PM = 0

WBJEE Previous Year (2010) - Question 7

The light beams of intensities in the ratio of 9 : 1 are allowed to interfere. What will be the ratio of the intensities of maxima and minima ?

Detailed Solution for WBJEE Previous Year (2010) - Question 7


WBJEE Previous Year (2010) - Question 8

If x1 be the size of the magnified image and x2 the size of the diminished image in Lens Displacement Method, then the size of the object is :

WBJEE Previous Year (2010) - Question 9

A point charge +q is placed at the centre of a cube of side L. The electric flux emerging from the cube is

Detailed Solution for WBJEE Previous Year (2010) - Question 9
Concept: Gauss Law. By Gauss theorem ... Total electric flux = Total charge inside cube /€. Flux = q/€. (where € is permittivity). Hence answer is 'A'.
WBJEE Previous Year (2010) - Question 10

Two capacitors of equal capacity are first connected in parallel and then in series. The ratio of the total capacities in the two cases will be

WBJEE Previous Year (2010) - Question 11

n identical droplets are charged to v volt each. If they coalesce to form a single drop, then its potential will be

Detailed Solution for WBJEE Previous Year (2010) - Question 11

 

WBJEE Previous Year (2010) - Question 12

The reading of the ammeter in the following figure will be 

Detailed Solution for WBJEE Previous Year (2010) - Question 12


WBJEE Previous Year (2010) - Question 13

A wire of resistance R is elongated n-fold to make a new uniform wire. The resistance of new wire

Detailed Solution for WBJEE Previous Year (2010) - Question 13

R' = n2R

WBJEE Previous Year (2010) - Question 14

The ratio of magnetic field and magnetic moment at the centre of a current carrying circular loop is x. When both the current and radius is doubled the ratio will be

Detailed Solution for WBJEE Previous Year (2010) - Question 14


WBJEE Previous Year (2010) - Question 15

The current through a coil of self inductance L = 2mH is given by I = t2 e–t at time t. How long it will take to make the e.m.f. zero?

Detailed Solution for WBJEE Previous Year (2010) - Question 15

I = t2 e–t

WBJEE Previous Year (2010) - Question 16

The magnetic flux through a loop of resistance 10 Ω is given by φ = 5t2 – 4t + 1 Weber. How much current is induced in the loop after 0.2 sec ?

Detailed Solution for WBJEE Previous Year (2010) - Question 16



WBJEE Previous Year (2010) - Question 17

The decimal equivalent of the binary number (11010.101)2 is

Detailed Solution for WBJEE Previous Year (2010) - Question 17

(11010.101) = 0 × 2º + 1 × 21 + 0 × 22 + 1 × 23 + 1 × 24 + 1 × 2–1 + 0 × 2–2 + 1 × 2–3 = 2 + 8 + 16 

WBJEE Previous Year (2010) - Question 18

In a common emitter configuration, a transistor has β = 50 and input resistance 1 kΩ. If the peak value of a.c. input is 0.01 V then the peak value of collector current is

Detailed Solution for WBJEE Previous Year (2010) - Question 18


WBJEE Previous Year (2010) - Question 19

Half-life of a radioactive substance is 20 minute. The time between 20% and 80% decay will be :

Detailed Solution for WBJEE Previous Year (2010) - Question 19

For 20% decay

For 80% decay

On dividing

WBJEE Previous Year (2010) - Question 20

The energy released by the fission of one uranium atom is 200 MeV. The number of fissions per second required to produce 3.2 W of power is (Take 1 eV = 1.6 × 10–19 J)

Detailed Solution for WBJEE Previous Year (2010) - Question 20

u = 200 MeV = 200 × 106 eV = 200 × 106 × 1.6 × 10–19 J
E = 3.2 J
No of fissions 

WBJEE Previous Year (2010) - Question 21

A body is projected with a speed u m/s at an angle β with the horizontal. The kinetic energy at the highest point is 3/4th of the initial kinetic energy. The value of β is :

Detailed Solution for WBJEE Previous Year (2010) - Question 21


K.E. = K cos2 β

WBJEE Previous Year (2010) - Question 22

A ball is projected horizontally with a velocity of 5 m/s from the top of a building 19.6 m high. How long will the ball take of hit the ground ?

Detailed Solution for WBJEE Previous Year (2010) - Question 22


WBJEE Previous Year (2010) - Question 23

A stone falls freely from rest and the total distance covered by it in the last second of its motion equals the distance covered by it in the first three seconds of its motion. The stone remains in the air for

Detailed Solution for WBJEE Previous Year (2010) - Question 23

u = 0

WBJEE Previous Year (2010) - Question 24

Two blocks of 2 kg and 1 kg are in contact on a frictionless table. If a force of 3 N is applied on 2 kg block, then the force ofcontact between the two blocks will be :

Detailed Solution for WBJEE Previous Year (2010) - Question 24

WBJEE Previous Year (2010) - Question 25

If momentum is increased by 20%, then kinetic energy increases by

Detailed Solution for WBJEE Previous Year (2010) - Question 25

WBJEE Previous Year (2010) - Question 26

A boy of mass 40 kg is climbing a vertical pole at a constant speed. If the coefficient of friction between his palms and the poleis 0.8 and g = 10 m/s2, the horizontal force that he is applying on the pole is

Detailed Solution for WBJEE Previous Year (2010) - Question 26

WBJEE Previous Year (2010) - Question 27

The value of ‘λ’ for which the two vectors a =5iˆ+λ jˆ+kˆ  and b =iˆ−2jˆ+kˆare perpendicular to each other is

Detailed Solution for WBJEE Previous Year (2010) - Question 27

WBJEE Previous Year (2010) - Question 28

If , then the angle included between and is

Detailed Solution for WBJEE Previous Year (2010) - Question 28

WBJEE Previous Year (2010) - Question 29

The height vertically above the earth’s surface at which the acceleration due to gravity becomes 1% of its value at the surface is (R is the radius of the Earth)​

Detailed Solution for WBJEE Previous Year (2010) - Question 29

WBJEE Previous Year (2010) - Question 30

The change in the gravitational potential energy when a body of mass m is raised to a height nR above the surface of the Earth is (here R is the radius of the Earth)

Detailed Solution for WBJEE Previous Year (2010) - Question 30

View more questions
3 videos|10 docs|54 tests
Information about WBJEE Previous Year (2010) Page
In this test you can find the Exam questions for WBJEE Previous Year (2010) solved & explained in the simplest way possible. Besides giving Questions and answers for WBJEE Previous Year (2010), EduRev gives you an ample number of Online tests for practice

Top Courses for JEE

Download as PDF

Top Courses for JEE