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Correct Answer is: (c, d)

The two particles have equal charges and move through the same p.d. They will, therefore, acquire the same energy (E). Also, E = p ²/ 2m or p = √(2mE), where p = momentum. For the same E, the particle with greater m has greater p.

Correct Answer is: (d) downwards

Correct Answer is: (d) if AB is given a small vertical displacement, it will undergo oscillations

The weight of AB is balanced by the ampere force on it. If it is pushed down a little, the ampere force increases. Hence, its acceleration is opposite to its displacement and it will undergo oscillation. 

Correct Answer is: (a, b, & d)

(b) 72 cm²

Here s_m = \(\frac{9+12+15}{2}\) = 18 cm, where lengths of medians are m₁ = 9 cm, m₂ = 12 cm, m₃ = 15 cm.

∴ Area of triangle = \(\frac{4}{3}\sqrt{18(18-9)(18-12)(18-15)}\) cm²

= \(\frac{4}{3}\sqrt{18\times9\times6\times3}\) cm² = \(\frac{4}{3}\) x 9 x 6 cm² = 72 cm².

Perimeter of rectangular field

= 2 x length + 2 x breadth

= 2 x 60 + 2 x 40

= 120 + 80

= 200 m

Hence, wire required for 4 rounds

= 200 x 4

= 800 m

∴ Wire required for 4 rounds

= 800 m

(d) 476 m²

Area of the octagonal surface

= 2 × \(\frac12\) × (22m + 10m) × 8m + 22m × 10m 

= 32m × 8m + 220 m² 

= 256 m² + 220 m² = 476 m²

Perimeter of a square

= 4 x length of one side

= 4 x 120

= 480 m

So, in one round the distance can be covered is 480 m, 

hence in 12 rounds the distance can be covered is

= 480 x 12 

= 5760 m 

= 5000 m + 760 m 

∴ Sanju covers 5 km 760 m daily

Given the length of a rectangle l = 14 cm 

Breadth of a rectangle b = ? cm 

Perimeter of the rectangle = 2(l + b) = 3 times of length. 

2(14 + b) = 3 x 14 

Divide with 2 on both sides,

\(\frac{2(14+b)}{2}=\frac{3\times14}{2}\)

14 + b = 21

Subtract with 14 on both sides.

14 + b – 14 = 21 – 14

Breadth b = 7cm

Area of the rectangle A = l x b = 14 x 7 = 98 sq.cm

Given :Length of the rectangle = 20 cm

Breadth of the rectangle = 15 cm

Base of the triangle = 30 cm

Area of the rectangle = Area of the triangle

Length x breadth = 1/2 × base × height

20 × 15 = 12 × 30 × height

∴ height = \(\frac {20\times15}{15} = 20 cm\)

(b) 26 units

From the question it is given that, area of rectangle = 36 units²

36 can be written as = 6 × 6

= (2 × 3) × (2 × 3)

= 2² × 3²

= 4 × 9

Then, the sides of the rectangle are 4 cm and 9 cm.

We know that, perimeter of the rectangle = 2 (length + breadth)

= 2 (4 + 9)

= 2 (13)

= 26 units

length of the empty space = 4 – 1 = 3 cm

breadth of the empty space = 3 – 1 = 2 cm 

square in empty space 

= length x breadth 

= 3 x 2 

= 6 sq.cm.

∴ 6 squares will fit in the empty space in the figure

Perimeter

= 2 + 1 + 1 + 2 + 1 + 1 + 2 + 1 + 1 + 2 + 1 + 1

= 16 cm 

∴ 16 cm

Perimeter of a rectangle 

= 2 x length + 2 x breadth 

= 2 x 10 + 2 x 8 

= 20 + 16 

= 36 cm ……… (i)

Perimeter of a square 

= 4 x length of side 

= 4 x 9 

36 cm …… (ii) 

From (i) and (ii) perimeter of both is equal. 

∴ perimeter of both is equal

The perimeter of a square room

= 4 × 1

= 4 × 15

= 60 meter

Unit square is the area of a unit square.

(1) The perimeter o f a square = 4 × 1 = 4 × 3 = 12 cm

(2) The perimeter of a square = 4 × 1 = 4 × 5 = 20 cm

(3) The perimeter of a square = 4 × 1 = 4 × 11 = 44 cm

(4) The perimeter of a square = 4 × 18 = 72 cm

(5) The perimeter of a square = 4 × 25 = 100 cm

(6) The perimeter of a square = 4 × 30 = 120 cm

(7) The perimeter of a square = 4 × 41 = 164 cm

(8) The perimeter of a square = 4 × 55 = 220 cm

(9) The perimeter of a square = 4 × 63 = 252 cm

(10) The perimeter of a square = 4 × 92 = 368 cm

(1) Area of Rectangle

= 1 × b

= 2 × 3

= 6 sq.cm

(2) Area of Rectangle

= 1 × b

= 2 × 4

= 8 sq.cm

(3) Area of Rectangle

= 1 × b

= 3 × 4

= 12 sq.cm

(4) Area of Rectangle

= 1 × b

= 5 × 4

= 20 Sq.cm

(5) Area of Rectangle

= 1 × b

= 5 × 2

= 10 Sq.cm

(6) Area of Rectangle

= 1 × b

= 3 × 6

= 18 Sq.cm

(7) Area of Rectangle

= 1 × b

= 4 × 6
= 24 Sq.cm

(8) Area ofRectangle

= 1 × b

= 5 × 6 = 30 Sq.cm

(9) Area of Rectangle

= 1 × b

= 8 × 6

= 48 Sq.cm

(10) Area of Rectangle

= 1 × b

= 7 × 9

= 63 Sq.cm

(1) The Perimeter of Rectangle = (21 + 2b)
= (2 × 2 + 2 × 3)
= 4 + 6
= 10 cm

(2) The Perimeter of Rectangle = (21 + 2b)

= (2 × 2 + 2 × 4)

= 4 + 8

= 12 cm

(3) The Perimeter of Rectangle = (21 + 2b)

= (2 × 3 + 2 × 4)

= 6 + 8

= 14 cm

(4) The Perimeter of Rectangle = (21 + 2b)

= (2 × 5 + 2 × 4)

= 10 + 8

= 18 cm

(5) The Perimeter of Rectangle = (21 + 2b)

= (2× 5 + 2 × 2)

= 10 + 4

= 14 cm

(6) The Perimeter of Rectangle = (21 + 2b)

= (2 × 3 + 2 × 6)

= 6 + 12

= 18 cm

(7) The Perimeter of Rectangle = (21 + 2b)

= (2 × 4 + 2 × 6)

= 8 + 12

= 20 cm

(8) The Perimeter of Rectangle = (21 + 2b)

= (2 × 5 + 2 × 6)

= (10 + 12)

= 22 cm

(9) The Perimeter of Rectangle = (21 + 2b)

= (2 × 8 + 2 × 6)

= 16 + 12

= 28 cm

(10) The Perimeter of Rectangle = (21 + 2b)

= (2 × 7 + 2 × 9)

= 14 + 18

= 32 cm

Side of squared field of Gopi = 75 m 

and area = 75 × 75 = 5625 sq. m 

and perimeter = 4 × side = 4 × 75 = 300 m 

Length of rectangular field of Narayan 85 m 

and b = a m (let) then

 Area = 85 × a = 85a sq. m and 

Perimeter= 2 (l + b) = 2 (85 + a) m 

According to question, both perimeter are equal 

∴ 2 (85 + a) = 300 

⇒ (85 + a) = 300/2 = 150 

⇒ a = 150 – 85 = 65 

Thus, breadth of field = 65 m then, 

area = 85 × 65 = 5525 m 

∵ 5625 > 5525 and 5625 – 5525 = 100 

∴ Area of Gopi’s field is more by 100 sq. m

(c) Difference of import and export

One of the following is not fossil fuel wood.

(c) Wind energy

Mechanical energy i.e. (K.E + P.E), heat energy, chemical energy, atomic energy, wind energy.

(c) Electric energy

Millions of years back when plants vegetation and animals present at that movement were supressed in the earth due to the excessive pressure and temperature is known as fossil fuel. Example are coal, petroleum, gas etc.

The energy which a body possess due to its position or due to change in the shape is called potential energy.
Example 1: A brick which is suspended above the ground has energy because it would do work by falling to the ground. Its energy is due to its position and therefore it is potential energy.
Example 2: The potential energy of a stretched bow string results from a change in its shape.

Wind mill is a machine which convert wind energy into mechanical energy. Wind mills are used for grinding of wheat and lifting water from the well.

The excreta of animals, cow dug, wild plants and dry leaves of plants are fermented in a digestive tank. The produced gas is called bio gas/Gobar gas which is known as bio mass energy. This gas bums with smokless flame. Therefore it is a non polluting gas. This is used as a house fuel and for lighting the house.

(b) Ocean energy

Water is held inside the dropper against the atmospheric pressure. When the rubber bulb is pressed, pressure on water becomes greater than the atmospheric pressure and so water comes out.

Due to the surface tension, the free surface of water behaves like stretched elastic membrane which is able to support the small weight of insects and they can move on the surface of water.

This is because the surface tension of soap solution is small as compared to the surface tension of water.

Correct Option is (C) 100 dynes

l = 5 cm

\(T_1 = 70\) dyne / cm

\(T = \frac{F}{l}\)

\(F_1 = T . l\)

\(F_1 = 5 \times 70\)

\(F_1 = 350\) dyne ............. \((F_1)\)

\(F_2 = T_2 .l\)

\(F_2 = 50 \times 5\)

\(F_2 = 250\) dyne .............. \((F_2)\)

Net. force = \(F_1 -F_2\) 

= 350 - 250

= 100 dynes

(C) 100 dynes

This is because the surface tension of soap solution is much smaller than that of pure water.

When we try to make a vertical film of pure water, it breaks due to the high surface tension of water.

The furious storm made the mast bend and the prow dip. The roaring storm chased the ship like a dreadful enemy. The ship sailed southwards. And there it had to face a very hostile weather. Now ‘there came both mist and snow’. Green ice as high as the mast came floating by the ship. The sailors had reached the land of mist and snow.

Oxygen and Nitrogen molecules.

Because the angle of refraction of different colours is different while passing through the glass prism.

Scattering of light takes place because of the particles present in the atmosphere. At high altitude due to the absence of atmospheric scattering of light do not take place and hence sky appears dark to passengers flying at high altitude.

If the earth has no atmosphere, no particles present either. Thus no scattering of light. Then, the sky appears dark.

• On a hot day, due to rise in the temperature, water vapour enters atmosphere which leads to abundant presence of water molecules in atmosphere.
• These water molecules scatter the colours of other frequencies (other than blue). 
• All such colours of other frequencies reach our eye and the sky appears white.

At very high altitude, there is no atmosphere. So there is no scattering of light at such heights. So sky appears dark to passengers.

Free surface of a liquid at rest behaves just like a stretched elastic membrane.

Since there is no water under the block to exert an upward force on it, therefore, there is no buoyant force.

(D) Both (A) and (B).

(D) the same in all three.

(A) cross-sectional area of the hole

(C) gauge pressure

Volume of the ball(V) = m/ρ

Mass of the liquid displaced by the ball,

m' = (m/ρ)ρ₀

When the ball falls with a constant velocity, then,

Viscous force = effective weight of the ball

or F = mg - m'g = (m - m')g

= (m -(mρ⁰/ρ))g = mg(1 - ρ₀/ρ)