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Physics MCQs for NEET — Practice Questions with Answers

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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 ?

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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$

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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

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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$

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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

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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 $ $= 16,000+20,000+9,000 = 45,000 cal$ $$= give by m_s mass of steam = ms L$$ $$ ms = { 45,000 \over 540 } = 83.3 g$$ $$convert into waters of 100 ^\circ C $$ $$Total water = 200 + 200 + 83.3 = 483.3 g$$ $$steam left = 100 - 83.3 = 16.79$$

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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 ?

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Explanation

$ W = Fd cos \theta $

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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 ?

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Explanation

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

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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) $

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Explanation

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

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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) $

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Explanation

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

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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...

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Explanation

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

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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) ] $

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Explanation

$ h_n \alpha (2n-1) $

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A spring of spring constant $ 10 ^3 n/m $ is stretched initially 4cm from the unstretched position. How much the work required to stretched it further by another 5 cm ?

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Explanation

$ W = { 1 \over 2} k ( x_2 ^2 -x_1^2 ) $

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