JEE Exam  >  JEE Tests  >  Test: BITSAT Past Year Paper- 2016 - JEE MCQ

Test: BITSAT Past Year Paper- 2016 - JEE MCQ


Test Description

30 Questions MCQ Test - Test: BITSAT Past Year Paper- 2016

Test: BITSAT Past Year Paper- 2016 for JEE 2024 is part of JEE preparation. The Test: BITSAT Past Year Paper- 2016 questions and answers have been prepared according to the JEE exam syllabus.The Test: BITSAT Past Year Paper- 2016 MCQs are made for JEE 2024 Exam. Find important definitions, questions, notes, meanings, examples, exercises, MCQs and online tests for Test: BITSAT Past Year Paper- 2016 below.
Solutions of Test: BITSAT Past Year Paper- 2016 questions in English are available as part of our course for JEE & Test: BITSAT Past Year Paper- 2016 solutions in Hindi for JEE course. Download more important topics, notes, lectures and mock test series for JEE Exam by signing up for free. Attempt Test: BITSAT Past Year Paper- 2016 | 150 questions in 180 minutes | Mock test for JEE preparation | Free important questions MCQ to study for JEE Exam | Download free PDF with solutions
Test: BITSAT Past Year Paper- 2016 - Question 1

What should be the velocity of rotation of earth due to rotation about its own axis so that the weight of a person becomes 3/5 of the presentweight at the equator. Equatorial radius of the earth is 6400 km.

Detailed Solution for Test: BITSAT Past Year Paper- 2016 - Question 1

True weight at equator, W = mg
Observed weight at equator,
W ' = mg ' = 3/5 mg
At equator, latitude λ = 0 
Using the formula,
mg ' = mg - mRω2 cos2λ
3/5 mg = mg - mRω2 cos20 = mg - mRω
⇒ mRω = mg - 3/5 mg = 2/5 mg

Test: BITSAT Past Year Paper- 2016 - Question 2

Block A of mass m and block B of mass 2m are placed on a fixed triangular wedge by means of a massless, inextensible string and a frictionless pulley as shown in figure. 


The wedge is inclined at 45° to the horizontal on both the sides. If the coefficient of friction between the block A and the wedge is 2/3 and that between the block B and the wedge is 1/3 and both the blocks A and B are released from rest, the acceleration of A will be

Detailed Solution for Test: BITSAT Past Year Paper- 2016 - Question 2



1 Crore+ students have signed up on EduRev. Have you? Download the App
Test: BITSAT Past Year Paper- 2016 - Question 3

The surface charge density of a thin charged disc of radius σ is s. The value of the electric field at the centre of the disc is σ/2∈0. With respect to the field at the centre, the electric field along the axis at a distance R from the centre of the disc

Detailed Solution for Test: BITSAT Past Year Paper- 2016 - Question 3

Test: BITSAT Past Year Paper- 2016 - Question 4

The molecules of a given mass of a gas have r.m.s. velocity of 200 ms–1 at 27°C and 1.0 × 105 Nm–2 pressure. When the temperature and pressure of the gas are respectively, 127°C and 0.05 × 105 Nm–2, the r.m.s. velocity of its molecules in ms–1 is :

Detailed Solution for Test: BITSAT Past Year Paper- 2016 - Question 4

Here v1 = 200 m/s;
temperature T1 = 27°C = 27 + 273 = 300 k
temperature T2 = 127° C = 127 + 273 = 400 k, V = ?
R.M.S. Velocity, V ∝ √T

Test: BITSAT Past Year Paper- 2016 - Question 5

An inductor of inductance L = 400 mH and resistors of resistance R1 = 2Ω and R2 = 2Ω are connected to a battery of emf 12 V as shown in the figure. The internal resistance of the battery is negligible. The switch S is closed at t = 0. The potential drop across L as a function of time is

Detailed Solution for Test: BITSAT Past Year Paper- 2016 - Question 5

Growth in current in LR2 branch when switch is closed is given by

Hence, potential drop across

Test: BITSAT Past Year Paper- 2016 - Question 6

Two wires are made of the same material and have the same volume. However wire 1 has crosssectional area A and wire 2 has cross-sectional area 3A. If the length of wire 1 increases by Δx on applying force F, how much force is needed to stretch wire 2 by the same amount?

Detailed Solution for Test: BITSAT Past Year Paper- 2016 - Question 6


As shown in the figure, the wires will have the same Young’s modulus (same material) and the length of the wire of area of crosssection 3A will be  ℓ/3 (same volume as wire 1).
For wire 1,

For wire 2,

From (i) and (ii),

⇒ F ' = 9F

Test: BITSAT Past Year Paper- 2016 - Question 7

Two spheres of different materials one with double the radius and one-fourth wall thickness of the other are filled with ice. If the time taken for complete melting of ice in the larger sphere is 25 minute and for smaller one is 16 minute, the ratio of thermal conductivities of the materials of larger spheres to that of smaller sphere is

Detailed Solution for Test: BITSAT Past Year Paper- 2016 - Question 7

Radius of small sphere = r Thickness of small sphere = t Radius of bigger sphere = 2r Thickness of bigger sphere = t/4 Mass of ice melted = (volume of sphere) × (density of ice)
Let K1 and K2 be the thermal conductivities of larger and smaller sphere.
For bigger sphere,

For smaller sphere,

Test: BITSAT Past Year Paper- 2016 - Question 8

A biconvex lens h as a radius of curvature of magnitude 20 cm. Which one of the following options best describe the image formed of an object of height 2 cm placed 30 cm from the lens?

Test: BITSAT Past Year Paper- 2016 - Question 9

In the figure below, what is the potential difference between the point A and B and between B and C respectively in steady state

Detailed Solution for Test: BITSAT Past Year Paper- 2016 - Question 9

The equivalent circuit is shown in figure.
V1 + V2 = 100
and   2V1 = 6V2

On solving above equations, we get
V1 = 75V, V2 = 25V

Test: BITSAT Past Year Paper- 2016 - Question 10

A radioactive element X converts into another stable element Y. Half life of X is 2 hrs. Initially only X is present. After time t, the ratio of atoms of X and Y is found to be 1 : 4, then t in hours is

Detailed Solution for Test: BITSAT Past Year Paper- 2016 - Question 10

Let N0 be the number of atoms of X at time t = 0.
Then at t = 4 hrs (two half lives)

∴ Nx/Ny = 1/3
and at t = 6 hrs  (three half lives)


The given ratio 1/4 lies between 1/3 and 1/7.
Therefore, t lies between 4 hrs and 6 hrs.

Test: BITSAT Past Year Paper- 2016 - Question 11

The approximate depth of an ocean is 2700 m. The compressibility of water is 45.4 × 10–11 Pa–1 and density of water is 103 kg/m3.What fractional compression of water will be obtained at the bottom of the ocean?

Detailed Solution for Test: BITSAT Past Year Paper- 2016 - Question 11

Compressibility of water,
K = 45.4 × 10–11 Pa–1
density of water P = 103 kg/m3
depth of ocean, h = 2700 m
We have to find ΔV/V = ?
As we know, compressibility,

So,(ΔV/V) = KPgh
= 45.4 × 10–11 × 103 × 10 × 2700
= 1.2258 × 10–2

Test: BITSAT Past Year Paper- 2016 - Question 12

A frictionless wire AB is fixed  on a sphere of radius R. A very small spherical ball slips on this wire. The time taken by this ball to slip from A to B is

Detailed Solution for Test: BITSAT Past Year Paper- 2016 - Question 12

Acceleration of body along AB is g cos q Distance travelled in time t sec =
From ΔABC, AB = 2R cos θ

Test: BITSAT Past Year Paper- 2016 - Question 13

A string of length ℓ is fixed at both ends. It is vibrating in its 3rd overtone with maximum amplitude 'a'.  The amplitude at a distance ℓ/3 from one end is

Detailed Solution for Test: BITSAT Past Year Paper- 2016 - Question 13

For a string vibrating in its nth overtone (n + 1)th harmonic)


Test: BITSAT Past Year Paper- 2016 - Question 14

A deuteron of kinetic energy 50 keV is describing a circular orbit of radius 0.5 metre in a plane perpendicular to the magnetic field B. The kinetic energy of the proton that describes a circular orbit of radius 0.5 metre in the same plane with the same B is

Test: BITSAT Past Year Paper- 2016 - Question 15

In the circuit shown in the figure, find the current in 45 Ω.

Detailed Solution for Test: BITSAT Past Year Paper- 2016 - Question 15


Test: BITSAT Past Year Paper- 2016 - Question 16

Kepler's third law states that square of period of revolution (T) of a planet around the sun, is proportional to third power of average distance r between sun and planet i.e. T2 = Kr3 here K is constant. If the masses of sun and planet are M and m respectively then as per Newton's law of gravitation force of attraction between them is here G is gravitational constant. The relation between G and K is described as

Detailed Solution for Test: BITSAT Past Year Paper- 2016 - Question 16

As we know, orbital speed, 
Time period

Squarring both sides,

Test: BITSAT Past Year Paper- 2016 - Question 17

Find the number of photon emitted per second by a 25 watt source of monochromatic light of wavelength 6600 Å. What is the photoelectric current assuming 3% efficiency for photoelectric effect ?

Detailed Solution for Test: BITSAT Past Year Paper- 2016 - Question 17

Pin = 25W, λ = 6600 Å = 6600 × 10–10 m nhv = P
⇒ Number of photons emitted/sec,

3% of emitted photons are producing current

Test: BITSAT Past Year Paper- 2016 - Question 18

A ray of light of intensity I is incident on a parallel glass slab at point A as shown in diagram. It undergoes partial reflection and refraction. At each reflection, 25% of incident energy is reflected. The rays AB and A'B' undergo interference. The ratio of Imax and Imin is :

Detailed Solution for Test: BITSAT Past Year Paper- 2016 - Question 18


Test: BITSAT Past Year Paper- 2016 - Question 19

A capillary tube of radius r is immersed vertically in a liquid such that liquid rises in it to height h (less than the length of the tube).  Mass of liquid in the capillary tube is m. If radius of the capillary tube is increased by 50%, then mass of liquid that will rise in the tube, is

Detailed Solution for Test: BITSAT Past Year Paper- 2016 - Question 19



New mass

Test: BITSAT Past Year Paper- 2016 - Question 20

The drift velocity of electrons in silver wire with cross-sectional area 3.14 × 10–6 m2 carrying a current of 20 A is. Given atomic weight of Ag = 108, density of silver = 10.5 × 103 kg/m3.

Detailed Solution for Test: BITSAT Past Year Paper- 2016 - Question 20

Number of electrons per kg of silver

Number of electrons per unit volume of silver

Test: BITSAT Past Year Paper- 2016 - Question 21

A parallel plate capacitor of area ‘A’ plate separation ‘d’ is filled with two dielectrics as shown. What is the capacitance of the arrangement?

Detailed Solution for Test: BITSAT Past Year Paper- 2016 - Question 21


(∵ C1 and C2 are in series and resultant of these two in parallel with C3)

Test: BITSAT Past Year Paper- 2016 - Question 22

In the Young’s double-slit experiment, the intensity of light at a point on the screen where the path difference is λ is K, (λ being the wave length of light used). The intensity at a point where the path difference is λ/4, will be :

Detailed Solution for Test: BITSAT Past Year Paper- 2016 - Question 22

For path differ ence λ, phase difference = 2π rad.
For path difference  λ/4, phase difference
= π/2 rad.
As K = 4I0  so intensity at given point where path difference is λ/4

Test: BITSAT Past Year Paper- 2016 - Question 23

The mass of 7N15 is 15.00011 amu, mass of 8O16 is 15.99492 amu and mp = 1.00783 amu. Determine binding energy of last proton of 8O16.

Detailed Solution for Test: BITSAT Past Year Paper- 2016 - Question 23

M(8O16) = M (7N15) + 1mP
binding energy of last proton
= M (N15) + mP – M (1O16)
= 15.00011 + 1.00783 – 15.99492
= 0.01302 amu = 12.13 MeV

Test: BITSAT Past Year Paper- 2016 - Question 24

A wire carrying current I has the shape as shown in adjoining figure. Linear parts of the wire are very long and parallel to X-axis while semicircular portion of radius R is lying in Y-Z plane. Magnetic field at point O is:

Test: BITSAT Past Year Paper- 2016 - Question 25

A stone projected with a velocity u at an angle θ with the horizontal reaches maximum height H1. When it is projected with velocity u at an angle with the horizontal, it reaches maximum height H2. The relation between the horizontal range R of the projectile, heights H1 and H2 is

Detailed Solution for Test: BITSAT Past Year Paper- 2016 - Question 25


and

Test: BITSAT Past Year Paper- 2016 - Question 26

If the series limit wavelength of Lyman series for the hydrogen atom is 912 Å then the series limit wavelength for Balmer series of hydrogen atoms is

Detailed Solution for Test: BITSAT Past Year Paper- 2016 - Question 26


For limiting wavelength of Lyman series

For limiting wavelength of Balmer series
n1 = 2, n2 = ∞

∴ λB = 4λL = 4 × 912 Å.

Test: BITSAT Past Year Paper- 2016 - Question 27

In the shown arrangement of the experiment of the meter bridge if AC corresponding to null deflection of galvanometer is x, what would be its value if the radius of the wire AB is doubled?

Detailed Solution for Test: BITSAT Past Year Paper- 2016 - Question 27


At null point 
If radius of the wire is doubled, then the resistance of AC will change and also the resistance of CB will change. But since R1/R2 does not change so, R3/R4 should also not change at null point. Therefore the point C does not change.

Test: BITSAT Past Year Paper- 2016 - Question 28

A 1 kg mass is attached to a spring of force constant 600 N/m and rests on a smooth horizontal surface with other end of the spring tied to wall as shown in figure. A second mass of 0.5 kg slides along the surface towards the first at 3m/s. If the masses make a perfectly inelastic collision, then find amplitude and time period of oscillation of combined mass.

Test: BITSAT Past Year Paper- 2016 - Question 29

The frequency of vibration of string is given by 

Here p is number of segments in the string and / is the length. The dimensional formula for m will be

Detailed Solution for Test: BITSAT Past Year Paper- 2016 - Question 29


Now, dimensional formula of R.H.S.

[p will have no dimension as it is an integer] = ML–1T0.
So, dimensions of m will be ML–1T0.

Test: BITSAT Past Year Paper- 2016 - Question 30

For the angle of minimum deviation of a prism to be equal to its refracting angle, the prism must be made of a material whose refractive index:

Detailed Solution for Test: BITSAT Past Year Paper- 2016 - Question 30


The angle of minimum deviation is given as
δmin = i + e–A
for minimum deviation
δmin = A then
 2A = i + e
in case of δmin i = e
2A = 2i r1 = r2 = A/2
i =  A = 90°
from smell’s law
1 sin i = n sin r1
sin A = n sin A/2


when  A = 90° = imin
then    nmin = √2
i = A = 0  nmax = 2

View more questions
Information about Test: BITSAT Past Year Paper- 2016 Page
In this test you can find the Exam questions for Test: BITSAT Past Year Paper- 2016 solved & explained in the simplest way possible. Besides giving Questions and answers for Test: BITSAT Past Year Paper- 2016, EduRev gives you an ample number of Online tests for practice

Top Courses for JEE

Download as PDF

Top Courses for JEE