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VITEEE Physics Test - 1


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35 Questions MCQ Test VITEEE: Subject Wise and Full Length MOCK Tests | VITEEE Physics Test - 1

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VITEEE Physics Test - 1 - Question 1

In a transformer 220 A.C. voltage is increased to 2200 volts. If the number of turns in the secondary are 2000, then the number of turns in the primary will be

Detailed Solution for VITEEE Physics Test - 1 - Question 1
Number of turns in primary coil ( Np ) = ? 

Number of turns in Secondary coil ( Ns ) = 2000 

Voltage in primary coil ( Ep ) = 220 V 

Voltage in secondary coil ( Es ) = 2200 V .

Now, using formula:  Ep/Es = Np/Ns

Substituting values, we get,

220/2200 = Np/2000 

=> Np = 200 turns . 
VITEEE Physics Test - 1 - Question 2

In a photo electric experiment the maximum velocity of photo electons emitted

VITEEE Physics Test - 1 - Question 3

Two wires of same metal have the same length but their cross sections are in the ratio 3:1. They are joined in series. The resistance of the thicker wire is 10 Ω. The total resistance of the combination is

Detailed Solution for VITEEE Physics Test - 1 - Question 3
VITEEE Physics Test - 1 - Question 4

If elements with principal quantum number n > 4 were not allowed in nature, then, the number of possible elements would be

Detailed Solution for VITEEE Physics Test - 1 - Question 4

Correct option is B)

Hint: 

Add all the elements from n=1 to n=4.

 

Step 1: Write the number of elements for each shell.

The number of elements for n=1 are 2.

The number of elements for n=2 are 8.

The number of elements for n=3 are 18.

The number of elements for n=4 are 32.

 

Step 2: adding all the elements,

from n=1 to n=4.

Total number =2+8+18+32=60

 

Hence option B correct

VITEEE Physics Test - 1 - Question 5

If a cell of constant electromotive force produces the same amount of heat during the same time in two independent resistances R₁ and R₂, when connected separately one after the other, across the cell, then the internal resistance of the cell is

Detailed Solution for VITEEE Physics Test - 1 - Question 5

 Answer :- d

Solution :- Here current through the resistance R1 is I1 = ε/(r+R1)

and current through the resistance R2 is I2 = ε/(r+R2).

Also, the heat produced due to passage of current through both the resistances is same.

Hence, Q = (I1)^2R1 = (I2)^2R2

Using the values of I1 and I2 in above equation we get (ε/(r+R1))^2 R1 = (ε/(r+R2))^2 R2 

where, ε is the emf and r is the internal resistance of battery.

ε^2 R1 / (r+R1)^2 = ε^2 R1 / (r+R1)^2

r^2(R1−R2) = R1R2((R1−R2​)

r^2= R1R2

∴ r = (R1R2)^1/2

VITEEE Physics Test - 1 - Question 6

Optical fibres uses the phenomenon of

VITEEE Physics Test - 1 - Question 7

A wire of radius r has resistance R. If it is stretched to a radius of 3r/4, its resistance becomes

VITEEE Physics Test - 1 - Question 8

The energy required to charge a parallel plate condenser of plate separation d and plate area of cross-section A such that the uniform electric field between the plates is E, is

Detailed Solution for VITEEE Physics Test - 1 - Question 8

Energy required = 1 2 CV2 = 1 2 ε0 E2Ad

VITEEE Physics Test - 1 - Question 9

Determine the current drawn from a 12 v supply with internal resistance 0.5Ω by the infinite network shown in the figure . Each resistor has Ω resistance

Detailed Solution for VITEEE Physics Test - 1 - Question 9

Let the equivalent resistance of the network be x×1

Rs=1+1+1+1

x = xΩ

The effective resistance x and 1Ω are in parallel

 

Now, the resistance Rp,1Ω and 1Ω are in series.

For infinite resistance

applying the quadratic formula

neglecting the negative value,

Resistance of the network

Total resistance of the circuit = 2.732 + 0.5 = 3.232Ω

Current 

VITEEE Physics Test - 1 - Question 10

A cell whose e.m.f. is 2V and internal resistance is 0.1Ω, is connected with a resistance of 3.9Ω. The voltage across the cell terminal will be

Detailed Solution for VITEEE Physics Test - 1 - Question 10

Answer :- c

Solution :- EMF of the cell(E) =2 V

internal resistance(r)= 0.1 V

External resistance(R)=3.9 V

V=IR

=(E÷(R+r))×R

=(2÷(3.9+0.1))×3.9

=1.95 V

Therefore potential difference across the terminals of the cell = 1.95V

VITEEE Physics Test - 1 - Question 11

In an electron gun, the electrons are accelerated by the potential V. If e is the charge and m is the mass of an electron, then the maximum velocity of these electrons will be

VITEEE Physics Test - 1 - Question 12

A bar magnet dropped into a coil of a conducting wire, along its axis, will fall with an acceleration

VITEEE Physics Test - 1 - Question 13

A generator at a utility company produces 100 A of current at 4000 V. The voltage is stepped up to 240000 V by a transformer before it is sent on a high voltage transmission line. The current in transmission line is

Detailed Solution for VITEEE Physics Test - 1 - Question 13

Given :     V1​=4000 volts      I1​=100 A              V2​=240000 volts

Using      V1​I1​=V2​I2​

∴     4000×100=240000×I2​              ⟹I2​=1.67 A

VITEEE Physics Test - 1 - Question 14
A 220 V, 1000 watt bulb is connected across a 110 V mains supply. The power consumed will be
VITEEE Physics Test - 1 - Question 15

Consider the situation in the figure. The work done in taking a point charge from P to Ais WA, from P to B is WB and from P to C is WC.
 

Detailed Solution for VITEEE Physics Test - 1 - Question 15

WA = WB = WC

Points A, B and C lie at the same distance from the charge q, i.e. they are lying on an equipotential surface. So, work done in moving a charge from A to B (WAB) or B toC (WBC) is zero.
Hence, work done in bringing a charge from P to A  = WA,
from P to B, WB = WA+WAB = WA
and from P to C, WC = WA + WAB + WBC = WA​

Hence, WA = WB = WC 

VITEEE Physics Test - 1 - Question 16

A dust particle of mass 10-3 gm is stationary between the plates of a horizontal parallel plate capacitor of 0 .016m separation which is connected to a voltage of 100V. How many fundamental charges (e = 1.6 x 10-19 C)   the dust particle carries

VITEEE Physics Test - 1 - Question 17

A capacitor of capacity C1 is charged upto V volt and then connected to an unchanged capacitor of capacity C2. The potential difference across each capacitor will be

Detailed Solution for VITEEE Physics Test - 1 - Question 17

Charge on first capacitor = q1 = C11
Charge on second capacitor = q2 = 0
When they are connected, in parallel the total charge
q = q1 + q2
∴ q = C11
and capacitance, C = C1 + C2
Let V' be the common potential difference across each capacitaor, then q = CV'
∴ V' = q C = C1V/ C1 + C2

VITEEE Physics Test - 1 - Question 18

A charge q is placed at the corner of a cube of side l. The electric flux passing through the cube is

Detailed Solution for VITEEE Physics Test - 1 - Question 18


Eight cubes, each side l, are required to form a Gaussian surface
So that the charge q at the corner of a small cube appears at the centre of the bigger cube
According to Gauss's law, the electric flux through the bigger cubes = q ∈ 0
Hence the electric flux through the given small cube = q 8∈ 0

VITEEE Physics Test - 1 - Question 19

Dielectric constant of a medium is unity. Its permittivity will be:

Detailed Solution for VITEEE Physics Test - 1 - Question 19

--

VITEEE Physics Test - 1 - Question 20

If a current is passed in a spring, it

Detailed Solution for VITEEE Physics Test - 1 - Question 20 Two conductors will attract each other if they have current flowing in same direction. in a spring each loop has current flowing in same direction resulting in contraction of the spring.
VITEEE Physics Test - 1 - Question 21

The phenomena in which proton flips is

VITEEE Physics Test - 1 - Question 22

The first atomic reactor was constructed by

VITEEE Physics Test - 1 - Question 23

The average binding energy of the nucleon is

VITEEE Physics Test - 1 - Question 24

If in a photoelectric experiment, the wavelength of incident radiation is reduced from 6000Å to 4000Å, then

Detailed Solution for VITEEE Physics Test - 1 - Question 24
Ans : (B)
ϕ=hcλ−eV∘ϕ= work function of a particular metal will remain constant
If λ = incident radiation wavelength is increased,
V∘ = stopping potential also has to increase to keep ϕ constant.
VITEEE Physics Test - 1 - Question 25

When a centimeter thick surface is illuminated with light of wavelength λ, stopping potential is V. When the same surface is illuminated by light of wavelength 2 λ, stopping potential is V/3. Threshold wavelength for metallic surface is

Detailed Solution for VITEEE Physics Test - 1 - Question 25
Let the threshold wavelength = λ
the energy required by photo electrons to escape = h c/λ
e V + hc/λ = hc/l
e V/3 + hc/λ = hc/2l
=>  h c/l = 4/3 * eV
=>  hc/λ = 1/3 * eV 
=>   λ = 4 l
VITEEE Physics Test - 1 - Question 26

In Ramsden's eye-piece, the distance from the eye lens at which the image due to objective of focal length f is formed is

Detailed Solution for VITEEE Physics Test - 1 - Question 26
The distance of image fanned by the objective lens from the eye lens = 2F/3 + F/4 
= 8F + 3F/12
= 11F/12
VITEEE Physics Test - 1 - Question 27

Which of the following spectrum have all the frequencies from high to low frequency range ?

VITEEE Physics Test - 1 - Question 28

While a collector emitter voltage is constant in a transistor, the collector current changes by 8.2 m A .When the emitter current changes by 8.3 m A, the value of forward current ratio hfe is

Detailed Solution for VITEEE Physics Test - 1 - Question 28
VITEEE Physics Test - 1 - Question 29

When the p-end of the p-n junction is connected to the negative terminal of the battery and the n-end to the positive terminal of the battery, then the p-n junction behaves like

VITEEE Physics Test - 1 - Question 30

Potential barrier developed in a junction diode opposes

VITEEE Physics Test - 1 - Question 31

In semiconductors, at the room temperature

VITEEE Physics Test - 1 - Question 32

In a semiconductor the separation between conduction band and valence band is of the order of

VITEEE Physics Test - 1 - Question 33

When the electrical conductivity of a semiconductor is due to the breaking of its covalent bonds, then the semiconductor is said to be

VITEEE Physics Test - 1 - Question 34

P-N junction diode is reverse biased when

VITEEE Physics Test - 1 - Question 35

Huygen's wave theory of light can not explain

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