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NEET Minor Test - 3 - NEET MCQ


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30 Questions MCQ Test NEET Mock Test Series 2025 - NEET Minor Test - 3

NEET Minor Test - 3 for NEET 2024 is part of NEET Mock Test Series 2025 preparation. The NEET Minor Test - 3 questions and answers have been prepared according to the NEET exam syllabus.The NEET Minor Test - 3 MCQs are made for NEET 2024 Exam. Find important definitions, questions, notes, meanings, examples, exercises, MCQs and online tests for NEET Minor Test - 3 below.
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NEET Minor Test - 3 - Question 1

A spherical ball is dropped in a long column of a highly viscous liquid. The curve in the graph shown, which represents the speed of the ball (v) as a function of time (t) is

Detailed Solution for NEET Minor Test - 3 - Question 1
Initial speed of ball is zero and it finally attains terminal speed

NEET Minor Test - 3 - Question 2

Given below are two statements: One is labelled as Assertion (A) and the other is labelled as Reason (R).

Assertion (A): The stretching of a spring is determined by the shear modulus of the material of the spring.

Reason (R): A coil spring of copper has more tensile strength than a steel spring of same dimensions.

In the light of the above statements, choose the most appropriate answer from the options given below

Detailed Solution for NEET Minor Test - 3 - Question 2
It is true that stretching of spring is determined by shear modulus of the spring as when coil spring is stretched neither its length nor its volume changes, there is only change in its shape.

Tensile strength of steel is more than that of copper.

Hence Assertion is true and reason is false.

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NEET Minor Test - 3 - Question 3

A The terminal velocity of a copper ball of radius 5mm falling through a tank of oil at room temperature is 10cms−1. If the viscosity of oil at room temperature is 0.9kgm−1s−1, the viscous drag force is:

Detailed Solution for NEET Minor Test - 3 - Question 3
Given, r = 5mm = 5 × 10−3m Vt = 10cms−1 = 10 × 10−2m−1s

Viscous drag force

F = 6πηrVt

F = 6 × π × 0.9 × 5 × 10−3 × 10 × 10−2

F = 84.78 × 10−4

F = 8.478 × 10−3N

NEET Minor Test - 3 - Question 4

The velocity of a small ball of mass M and density d, when dropped in a container filled with glycerine becomes constant after some time. If the density of glycerine is d/2, then the viscous force acting on the ball will be

Detailed Solution for NEET Minor Test - 3 - Question 4
I. Let Fv be the viscous force and FB be the Bouyant force acting on the ball.

Then, when body moves with constant velocity

Mg = FB + Fv[a = 0]

Fv = Mg − FB

NEET Minor Test - 3 - Question 5

A capillary tube of radius r is immersed in water and water rises in it to a height h. The mass of the water in the capillary is 5g. Another capillary tube of radius 2r is immersed in water. The mass of water that will rise in this tube is :

Detailed Solution for NEET Minor Test - 3 - Question 5
Force of surface tension balances the weight of water in capillary tube.

FS = 2πrT cosθ = mg

Here, T and θ are constant.

So, m ∝ r

Let m1 and m2 be the mass of water in two capillary tube.

NEET Minor Test - 3 - Question 6

A small hole of area of cross-section 2mm2 is present near the bottom of a fully filled open tank of height 2m. Taking g = 10m/s2, the rate of flow of water through the open hole would be nearly

Detailed Solution for NEET Minor Test - 3 - Question 6
According to Torricelli's theorem, Velocity,

From equation of continuity, Volume of liquid flowing per second, Q = Av

= 2 × 10−6 × 6.32 = 12.6 × 10−6m3/s

NEET Minor Test - 3 - Question 7

A soap bubble, having radius of 1 mm, is blown from a detergent solution having a surface tension of 2.5 × 10−2N/m.

The pressure inside the bubble equals at a point Z0 below the free surface of water in a container. Taking g = 10m/s2, density of water = 103kg/m3, the value of Z0 is

Detailed Solution for NEET Minor Test - 3 - Question 7
The pressure at a point Z0 below the surface of water,

Also, pressure inside a soap bubble,

NEET Minor Test - 3 - Question 8

The stress-strain curves are drawn for two different materials X and Y. It is observed that the ultimate strength point and the fracture point are close to each other for material X but are far apart for material Y. We can say that materials X and Y are likely to be (respectively)

Detailed Solution for NEET Minor Test - 3 - Question 8
Fracture point and ultimate strength point is close for material X, hence X is brittle in nature and both points are far apart for material Y hence it is ductile.
NEET Minor Test - 3 - Question 9

Two wires are made of the same material and have the same volume. The first wire has cross-sectional area A and the second wire has cross-sectional area 3A. If the length of the first wire is increased by ∆l on applying a force F, how much force is needed to stretch the second wire by the same amount?

Detailed Solution for NEET Minor Test - 3 - Question 9
Young's modulus,

since initial volume of wires are same

∴ Their areas of cross sections are A and 3A and lengths are 3l and l respectively.

For wire 1

For wire 2, let F′ force is applied

NEET Minor Test - 3 - Question 10

Two rods A and B of different materials are welded together as shown in figure. Their thermal conductivities are K1 and K2 The thermal conductivity of the composite rod will be

Detailed Solution for NEET Minor Test - 3 - Question 10
Equivalent thermal conductivity of the composite rod in parallel combination will be,

NEET Minor Test - 3 - Question 11

A U tube with both ends open to the atmosphere, is partially filled with water. Oil, which is immiscible with water, is poured into one side until it stands at a distance of 10 mm above the water level on the other side. Meanwhile the water rises by 65 mm from its original level (see diagram). The density of the oil is

Detailed Solution for NEET Minor Test - 3 - Question 11
Pressure at point C,

Pc = Pa + ρwaterghwater

where hwater = CE = (62 + 65)mm = 130mm

Pressure at point B, PB = Pa + ρoil ghoul

where

hoil = AB = (65 + 65 + 10)mm = 140mm

In liquid, pressure is same at same level of liquid,

PB = PC

NEET Minor Test - 3 - Question 12

A rectangular film of liquid is extended from (4cm × 2cm) to (5cm × 4cm). If the work done is 3 × 10−4J, the value of the surface tension of the liquid is

Detailed Solution for NEET Minor Test - 3 - Question 12
Work done = Surface tension of film x Change in area of the film

or W = T × ΔA

Here, A1 = 4cm × 2cm = 8cm2

A − 2 = 5cm × 4cm = 20cm2

ΔA = 2(A2 − A1) = 24cm2 = 24 × 10−4m2

W = 3 × 10−4J,T = ?

NEET Minor Test - 3 - Question 13

Two identical bodies are made of a material for which the heat capacity increases with temperature. One of these is at 100°C, while the other one is at 0°C. If the two bodies are brought into contact, then, assuming no heat loss, the final common temperature is

Detailed Solution for NEET Minor Test - 3 - Question 13
Since, heat capacity of material increases with increase in temperature so, body at 100 °C has more heat capacity than body at 0°C. Hence, final common temperature of the system will be closer to 100\°C

∴ Tc > 50°C

NEET Minor Test - 3 - Question 14

Coefficient of linear expansion of brass and steel rods are α1 and α2. Length of brass and steel rods are l1 and l2 respectively. If (l2 − l1) is maintained same at all temperatures, which one of the following relations holds good?

Detailed Solution for NEET Minor Test - 3 - Question 14
Linear expansion of brass = α1

Linear expansion of steel = α2

Length of brass rod = l1

Length of steel rod = l2

On increasing the temperature of the rods by ΔT, new lengths would be

l1′ = l1(1 + α1ΔT) ........(i)

l2′ = l2(1 + α2ΔT) .........(ii)

Subtracting eqn. (i) from eqn. (ii), we get

l2′ − l1′ = (l2 − l1) + (l2α2 − l1α1)ΔT

According to question,

l2′ − l1′ = l2 − l1  (for all temperatures)

∴ l2α2 − l1α1 or l1α1 = l1α2

NEET Minor Test - 3 - Question 15

A black body is at a temperature of 5760 K. The energy of radiation emitted by the body at wavelength 250nm is U1, at wavelength 500 nm is U2 and that at 1000 nm is U3 Wien's constant, b = 2.88 × 106nmK. Which of the following is correct?

Detailed Solution for NEET Minor Test - 3 - Question 15
According to Wein's displacement law

Clearly from graph, U1 < />2 < />3

NEET Minor Test - 3 - Question 16

An airplane of a total wing area of 120 m2 is in a level flight at some height. If the pressure difference between the upper and lower surface is 2.5 kPa, then the mass of the airplane is: [Take g = 10 ms-2]

Detailed Solution for NEET Minor Test - 3 - Question 16
Step 1: Balance the forces

(P1 - P2)A = mg

Step 2: Calculate the mass

Thus, m = 2.5 × 103 × 12010 = 3 × 104 kg.

NEET Minor Test - 3 - Question 17

A liquid is poured into three vessels of the same base area and equal heights as shown in the figure, then:

Detailed Solution for NEET Minor Test - 3 - Question 17
Step 1: Find the pressure at the bottom in all cases

Pb = P0 + ρgh

Step 2: Find thrust at the base of each vessel

F = PbA

= PbA

As the pressure and the area is the same for all the cases. So, the force on the base will be equal for all the vessels.

NEET Minor Test - 3 - Question 18

If pressure at half the depth of a lake is equal to 2/3rd the pressure at the bottom of the lake, then the depth of the lake is:

Detailed Solution for NEET Minor Test - 3 - Question 18
Pressure at half the depth

Pressure at the bottom = P0 + hdg

According to the given condition

NEET Minor Test - 3 - Question 19

The value of g at a place decreases by 2%. Then, the barometric height of mercury:

Detailed Solution for NEET Minor Test - 3 - Question 19

If the value of g is decreased by 2% then h will increase by 2%.

NEET Minor Test - 3 - Question 20

A siphon in use is demonstrated in the following figure. The density of the liquid flowing in the siphon is 1.5 gm/cc. The pressure difference between the point P and S will be:

Detailed Solution for NEET Minor Test - 3 - Question 20
As both the points are at the surface of the liquid and in the open atmosphere. So, both the points possess similar pressure equal to 1 atm. Hence the pressure difference will be zero.
NEET Minor Test - 3 - Question 21

From the adjacent figure, the correct observation is:

Detailed Solution for NEET Minor Test - 3 - Question 21
Pressure = hρg i.e. pressure at the bottom is independent of the area of the bottom of the tank. It depends on the height upto which the tank is filled with water. As in both the tanks, the levels of water are the same, pressure at the bottom is also the same.
NEET Minor Test - 3 - Question 22

A diver is 10 m below the surface of water. The approximate pressure experienced by the diver is-

Detailed Solution for NEET Minor Test - 3 - Question 22
P = P0 + ρgh

Step 1: Put the values in above formula and find P.

= 105 + 103 × 10 × 10

= 2 × 105Pa

NEET Minor Test - 3 - Question 23

The area of cross-section of the wider tube shown in the figure is 800 cm2. If a mass of 12 kg is placed on the massless piston, then the difference in heights h of the levels of water in the two tubes will be:

Detailed Solution for NEET Minor Test - 3 - Question 23
Pressure should be same on both sides.

NEET Minor Test - 3 - Question 24

The height of a mercury barometer is 75 cm at sea level and 50 cm at the top of a hill. The ratio of density of mercury to that of air is 104. The height of the hill is:

Detailed Solution for NEET Minor Test - 3 - Question 24
Difference of pressure between sea level and top of hill

∆P = (h1 - h2) × ρHg × g = (75 - 50) × 10-2 × ρHg × g …(i)

and pressure difference due to h meter of air

∆P = h × ρair × g …(ii)

By equating (i) and (ii) we get

h × ρair × g = (75 − 50) × 10−2 × ρHg × g

Height of the hill = 2.5 km.

NEET Minor Test - 3 - Question 25

A vertical U-tube of uniform inner cross section, contains mercury in both its arms. A glycerin (density = 1.3 g/ cm3) column of length 10 cm is introduced into one of its arms. Oil of density 0.8 gm/ cm3 is poured into the other arm until the upper surfaces of the oil and glycerin are at the same horizontal level. Find the length of the oil column. [Density of mercury = 13.6 g/cm3]

Detailed Solution for NEET Minor Test - 3 - Question 25

At the condition of equilibrium

Pressure at point A = Pressure at point B

PA = PB ⇒ 10 × 1.3 × g = h × 0.8 × g + (10 - h) × 13.6 × g

By solving we get h = 9.7 cm

NEET Minor Test - 3 - Question 26

The value of the coefficient of volume expansion of glycerin is 5 x 10-4 K-1. The fractional change in the density of glycerin for a temperature increase of 40°C will be:

Detailed Solution for NEET Minor Test - 3 - Question 26

NEET Minor Test - 3 - Question 27

When a uniform rod is heated, which of its following properties will increase as a result of it?

Detailed Solution for NEET Minor Test - 3 - Question 27
Step 1:

As the rod is heated its length increases.

Step 2: As the moment of inertia of the rod depends on its length so, it will changes.

NEET Minor Test - 3 - Question 28

If two rods of length L and 2L having coefficients of linear expansion α and 2α respectively are connected so that their total length becomes 3L, the average coefficient of linear expansion of the composition of rods equals:

Detailed Solution for NEET Minor Test - 3 - Question 28

NEET Minor Test - 3 - Question 29

Two rods, one made of aluminium and the other made of steel, having initial lengths l1 and l2 are connected together to form a single rod of length l1 + l2. The coefficient of linear expansion for aluminium and steel are αa and αs respectively. If the length of each rod increases by the same amount when their temperature is raised by t°C, then the ratio

Detailed Solution for NEET Minor Test - 3 - Question 29
Step 1: Find the change in the lengths of the two rods.

We have,

Step 2: Find the correct answer.

NEET Minor Test - 3 - Question 30

A large steel wheel is to be fitted onto a shaft of the same material. At 27 °C, the outer diameter of the shaft is 8.70 cm and the wheel's central hole has a diameter of 8.69 cm. The shaft is cooled using ‘dry ice’. At what temperature of the shaft does the wheel slip on the shaft? (Assume the coefficient of linear expansion of the steel to be constant over the required temperature range and α steel = 1.20 × 10−5K−1)

Detailed Solution for NEET Minor Test - 3 - Question 30
The given temperature, T = 27° C = 27 + 273 = 300 K

Step 1: Use the formula of linear expansion.

We know that,

Δd = -d1αsteel(T1 - T)

T1 is temperature after cooling.

Step 2: Find the final temperature.

0.01 = -8.7 × 1.20 × 10-5 × (T1 - 300)

(T1 - 300) = -95.78

T1 = 204.21 K

= -68.95°C

Therefore the wheel will slip on the shaft when the temperature is -69 °C.

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