NEETshala
Log in Get Started Free

NEET Practice Questions (MCQs) with Answers & Solutions

Practice free NEET NEET multiple-choice questions online with instant answers and detailed explanations. No login required.

All Physics Chemistry Botany Zoology
Register free to filter questions
● ● ● ● ●

In Column I different Process is given match corresponding option of column II Column - I (A) adiabatic process (B) Isobaric process (C) Isochroic process (D) Isothermal process Column - II (p) $ \triangle p = 0 $ (q) $ \triangle u = 0 $ (.r) $ \triangle Q = 0 $ (s) $ \triangle W = 0 $

You've reached today's free limit of 20 questions. Log in to keep practising for free.
Explanation

adiabatic process $ \triangle Q = 0 $ Isobasic process P = const $ \therefore \triangle P = 0 $ Isochroic process V = const $ \therefore \triangle W = 0 $ Isothermal process T = const $ \therefore \triangle u = 0 $

● ● ● ● ●

In a containes of negligible heat capacity, 200g ice at $0 ^\circ C $ and 100g steam at $ 100 ^\circ C $ are added to 200g of water that has temperature $ 55 ^\circ C $ . Assume no heat is lost to the surroundings and the pressure in the container is constant 1 atm. What is the final temperature the System ?

You've reached today's free limit of 20 questions. Log in to keep practising for free.
Explanation

head required by ice and water to go up to $ 100^\circ C = m_1L+ m_1 sw \triangle T + mw sw \triangle T$ $ = 200 \times 80+200 \times 1 \times 100+200 \times 1 \times 45 $ $= 16,000+20,000+9,000 = 45,000 cal$ $ = give by m_s mass of steam = ms L$ $ ms = { 45,000 \over 540 } = 83.3 g$ $ convert into waters of 100 ^\circ C $ $ Total water = 200 + 200 + 83.3 = 483.3 g$ $ steam left = 100 - 83.3 = 16.79$

● ● ● ● ●

In a containes of negligible heat capacity, 200g ice at $0 ^\circ C $ and 100g steam at $ 100 ^\circ C $ are added to 200g of water that has temperature $ 55 ^\circ C $ . Assume no heat is lost to the surroundings and the pressure in the container is constant 1 atm. At the final temperature, mass of the total water present in the system is

You've reached today's free limit of 20 questions. Log in to keep practising for free.
Explanation

head required by ice and water to go up to $ 100^\circ C = m_1L+ m_1 sw \triangle T + mw sw \triangle T$ $ = 200 \times 80+200 \times 1 \times 100+200 \times 1 \times 45 $ $= 16,000+20,000+9,000= 45,000 cal$ $= give by m_s mass of steam = ms L$ $ ms = { 45,000 \over 540 } = 83.3 g$ $convert into waters of 100 ^\circ C $ $Total water = 200 + 200 + 83.3 = 483.3 g$ $ steam left = 100 - 83.3 = 16.79$

● ● ● ● ●

In a containes of negligible heat capacity, 200g ice at $0 ^\circ C $ and 100g steam at $ 100 ^\circ C $ are added to 200g of water that has temperature $ 55 ^\circ C $ . Assume no heat is lost to the surroundings and the pressure in the container is constant 1 atm. Amount of the Sm left in the system, is equal to

You've reached today's free limit of 20 questions. Log in to keep practising for free.
Explanation

head required by ice and water to go up to $ 100^\circ C = m_1L+ m_1 sw \triangle T + mw sw \triangle T$ $ = 200 \times 80+200 \times 1 \times 100+200 \times 1 \times 45 $ $= 16,000+20,000+9,000 = 45,000 cal$ $$= give by m_s mass of steam = ms L$$ $$ ms = { 45,000 \over 540 } = 83.3 g$$ $$convert into waters of 100 ^\circ C $$ $$Total water = 200 + 200 + 83.3 = 483.3 g$$ $$steam left = 100 - 83.3 = 16.79$$

● ● ● ● ●

How much is the work done in pulling up a block of wood weighing 2KN for a length of 10m on a smooth plane inclined at an angle of 30o with the horizontal ?

You've reached today's free limit of 20 questions. Log in to keep practising for free.
Explanation

$ W = Fd cos \theta $

● ● ● ● ●

A force of 7N, making an angle $ \theta $ with the horizantal, acting on an object displaces it by 0.5m along the horizontal direction. If the object gains K.E. of 2J, what is the horizontal component of the force ?

You've reached today's free limit of 20 questions. Log in to keep practising for free.
Explanation

$ \triangle K = Fd cos \theta = w $

● ● ● ● ●

A 60 kg JATAN with 10 kg load on his head climbs 25 steps of 0.20m height each. what is the work done in climbing ? $ (g = 10 m/s^2) $

You've reached today's free limit of 20 questions. Log in to keep practising for free.
Explanation

$ W = Fd = (mg) (d) ; where M = m_1 +m_2 $

● ● ● ● ●

A ball of mass 5 kg is stiding on a plane with intial velocity of 10 m/s. If co- efficient of friction between surface and ball is 1/2 then before stopping it will describe...... $ (g = 10 m/s^ 2) $

You've reached today's free limit of 20 questions. Log in to keep practising for free.
Explanation

$ a = \mu g and v^2 = v_0^2 = -2 \mu gd $ $ \therefore d = { vo^2 \over 2 \mu g } $ (v = 0)

● ● ● ● ●

The force constant of a wire is K and that of the another wire is 3k when both the wires are stretched through same distance, if work done are $W_1 and W_2,$ then...

You've reached today's free limit of 20 questions. Log in to keep practising for free.
Explanation

According $ W = { 1 \over 2 } kx^2 $

● ● ● ● ●

A ball is released from the top of a tower. what is the ratio of work done by force of gravity in first, second and third second of the motion of the ball ?
$ [ h_n \alpha (2n-1) ] $

You've reached today's free limit of 20 questions. Log in to keep practising for free.
Explanation

$ h_n \alpha (2n-1) $

Ready to ace NEET?

Free access · No credit card required

Browse All Questions Create Free Account

Frequently Asked Questions

Yes. You can attempt every NEET question on this page for free without logging in, and check the correct answer with a detailed explanation instantly.

No account is required to attempt questions and view answers. A free account adds bookmarks, personal notes, and progress tracking.

The bank mixes NEET previous year questions (PYQs) with practice questions, each tagged with its exam appearances where applicable.