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• Table salt, or common salt, which gives a salty taste to food, is the most used of all salts. 
• Its chemical name is Sodium chloride.
• Sodium chloride is formed by a neutralization reaction between sodium hydroxide and hydrochloric acid.
• Chemical reaction.
• This is a neutral salt and the pH value of its aqueous solution is 7.

Correct answer is: (a) 25

Correct answer is: (b) yellowish-white

Correct answer is: (c) brown

8 से विभाज्य संख्याएँ = 2496, 7000, 384, 560, 960

12 से विभाज्य संख्याएँ = 2496, 5700, 300, 384, 960

15 से विभाज्य संख्याएँ = 900, 6225, 3690

ट्रांजिस्टर का मूल्य = 1160 रुपए।

टी०वी० का मूल्य = 1160 + 7190 = 8350 रुपए

एक ट्रांजिस्टर और एक टी०वी० सेट का मूल्य = 1160 + 8350 = 9510 रुपए

अतः एक ट्रांजिस्टर और एक टी०वी० सेट का मूल्य = 9510

(घ) अभाज्य  

जिस संख्या का स्वयं और 1 के अतिरिक्त और कोई गुणनखण्ड नहीं होता, उसे अभाज्य कहते हैं

गुणक = 999, गुणनफल = 744255

गुण्य = गुणनफल ÷ गुणक = 744255 ÷ 999 = 745

105 किग्रा चावल भरा जाता है = 1 बोरे में

5775 किग्रा चावल भरा जाएगा = 5775 ÷ 105 = 55 बोरे

(क)

18 = 2 x 3 x 3

24 = 2 x 2 x 2 x 3

म०स० = 2 x 3 = 6

(ख)

4 = 2 x 2

8 = 2 x 2 x 2

12 = 2 x 2 x 3

म०स० = 2 x 2 = 4

1 मिनट = 1/60 घंटा, 1800 मिनट = 1800/60 = 30 घंटा

(क) चारों ओर जल से घिरे भूमि के छोटे भाग को द्वीप कहते हैं।

(ख) बंगाल की खाड़ी में स्थित द्वीपसमूह का नाम अंडमान-निकोबार द्वीप समूह है।

(ग) अरब सागर में स्थित द्वीपसमूह का नाम लक्षद्वीप समूह है।

(d) λ = 7, μ = -5

(c) consistent

(a) The rank of the matrix is 1 

(b) consistent

(a) no solution

(d) infinitely many solution

We know that,

\(\tan^{-1}\left(\frac{A+B}{1-AB}\right)\)\(=\tan^{-1}A+\tan^{-1}B\)

Now, taking A = x and B = √x 

We get,

\(\tan^{-1}\mathrm x+\tan ^{-1}\sqrt{\mathrm x}\)\(=\tan^{-1}\left(\frac{\mathrm x+\sqrt{\mathrm x}}{1-\mathrm x^{3/2}}\right)\)

As, x. x ^1/2 = x ^3/2 

Hence, Proved.

We know that,

\(\tan^{-1}\left(\frac{A+B}{1-AB}\right)\)\(=\tan^{-1}A+\tan^{-1}B\)

Now, taking A = x and B = √x 

We get,

\(\tan^{-1}\mathrm x+\tan^{-1}\sqrt{\mathrm x}\)\(=\tan^{-1}\left(\frac{\mathrm x+\sqrt{\mathrm x}}{1-\mathrm x^{3/2}}\right)\)

As, x. x ^1/2 = x ^3/2 

Hence, Proved.

To Prove:

\(\tan^{-1}\left(\frac{\sqrt{\mathrm x}+\sqrt{\mathrm y}}{1-\sqrt{\mathrm {xy}}}\right)=\tan^{-1}\sqrt{\mathrm x}+tan^{-1}\sqrt{\mathrm y}\)

We know that, 

\(\tan A+\tan B = \frac{\tan A+\tan B}{1-\tan A \tan B}\)

Also, \(\tan^{-1}\left(\frac{A+B}{1-AB}\right)=\tan^{-1}A + \tan ^{-1} B\)

Taking A = √x and B = √y 

We get,

\(\tan^{-1}\left(\frac{\sqrt{\mathrm x}+\sqrt{\mathrm y}}{1-\sqrt{\mathrm{xy}}}\right)\)\(=\tan^{-1}\sqrt{\mathrm x}+\tan^{-1}\sqrt{\mathrm y}\)

Hence, Proved.

(i) True A point whose x coordinate is zero and y-coordinate is non-zero will lie on the y-axis.

(ii) False The point whose y coordinate is zero and x-coordinate is 5 will lie on y-axis.

(iii) True The coordinates of the origin are (0, 0).

(i) Take shelter under a table and stay there till the shaking stops. 

(ii) Stay away from tall and heavy objects that may fall on you. 

(iii) If you are in bed, do not get up. Protect your head with a pillow. (If a child gives any other reasonable precautions, it should be accepted)

In order to protect ourselves during an earthquake we must protect ourselves in the following ways:

1. We must take shelter under the table and should stay until the shaking stops.

2. We must stay away from the heavy objects to avoid them to fall on us. 

3. If we are in bed we must stay in it and should protect our head with a pillow.

Data : p – p₀ = 51.2 Pa, T = 3.2 × 10^-2 N/m

Forasoapbubb1e, p – p₀ = \(\frac{4T}R\)

∴ The radius of the soap bubble should be

R = \(\frac{4T}{p-p_0}\) = \(\frac{4\times3.2\times10^{-2}}{51.2}\) = 2.5 × 10^-3 m = 2.5 mm

∴ the diameter of the soap bubble should be 2 × 2.5 = 5 mm.

The force acting on the body at t = 2 seconds is - (a) 136 N

Initial momentum to the ball A = 0.05(6) = 0.3 kg m/s 

As the speed is reversed on collision,

final momentum of ball A = 0.05(-6) = -0.3 kg m/s 

Impulse imparted to ball A = change in momentum of ball A = final momentum – initial momentum = -0.3 -0.3 = -0.6 kg m/s. 

(c) The impulse imparted to each ball is 0.5 Ns.

(d) The impulse and the force on each ball are equal in magnitude and opposite in direction.

Correct option is: (D) 500 m/s

Using conservation of linear momentum.

\(m_1 \ v_1 = m_2 \ v_2\)

\(v_1 = \frac{m_2 \ \times \ v_2}{m_1}\)

\(v_1 = \frac{1 \times 5}{10 \times 10^{-3}}\)

\(v_1 = \frac{5000}{10}\)

\(v_1 = 500 \ m/s\)

Correct option is: (D) 500 m/s

Mass of each ball = 0.05 kg

Initial velocity of each ball = 6 m/s

Magnitude of the initial momentum of each ball, pi = 0.3 kg m/s

After collision, the balls change their directions of motion without changing the magnitudes of their velocity. 

Final momentum of each ball, pf = –0.3 kg m/s 

Impulse imparted to each ball = Change in the momentum of the system

= pf – pi 

= –0.3 – 0.3 = –0.6 kg m/s 

The negative sign indicates that the impulses imparted to the balls are opposite in direction.

Let m₁ & m₂ be the masses of smaller nuclei and let \(\vec v_1\) & \(\vec v_2\) be their velocities.

According to the law of conservation of momentum.

Initial momentum = final momentum

0 = m₁ \(\vec v_1\) + \(\vec v_2\)

or \(\vec v_1\)= - \(\frac {-m_1}{m_2}\)\(\vec v_1\)

Hence v₁ & v₂ are in opposite direction.

Mass of each ball = 0.05 kg
Initial velocity of each ball = 6 m/s
Magnitude of the initial momentum of each ball, p_i = 0.3 kg m/s
After collision, the balls change their directions of motion without changing the magnitudes of their velocity.
Final momentum of each ball, p_f = –0.3 kg m/s
Impulse imparted to each ball = Change in the momentum of the system
= p_f – p_i
= –0.3 – 0.3 = –0.6 kg m/s
The negative sign indicates that the impulses imparted to the balls are opposite in direction.

Mass of the gun, M = 100 kg

Mass of the shell, m = 0.020 kg

Muzzle speed of the shell, v = 80 m/s

Recoil speed of the gun = V

Both the gun and the shell are at rest initially.

Initial momentum of the system = 0 

Final momentum of the system = mv – MV

Here, the negative sign appears because the directions of the shell and the gun are opposite to each other.

According to the law of conservation of momentum: 

Final momentum = Initial momentum

mv – MV = 0

V = mv/M

= 0.020 x /100 x 1000 = 0.016 m/s

Impulse = change in momentum

Initial momentum of each ball = 0.05 × 6

= 0.3 kg m/s

Final momentum of each ball = 0.05 × (-6)

= – 0.3 kg m/s

Impulse = 0.6 kgm/s (in magnitude).

m₁ = 0.02 kg, m₂ = 100 kg

v₁ = 80 m/s v₂ = ?

According law of conservation of momentum

m₁u₁ + m₂u₂ = m₁v₁ + m₂v₂

0.02 (0) + 100 (0) = 0.02 × 80 + 100 × v₂

v₂ = – 1.6 × 10^-2 m/s.

Given: m₁ = 0.02 kg, m₂ = 100 kg, v₁ = 80 m s^-1

To find: Recoil speed (v₂)

Formula: m₁u₁ + m₂u₂ = m₁v₁ + m₂v₂

Calculation: Initially gun and shell are at rest.

∴ m₁u₁ + m₂u₂ = 0

Final momentum = m₁v₁ – m₂v₂

Using formula,

0 = 0.02 (80) – 100(v₂)

∴ v₂ = \(\frac{0.02\times80}{100}\) = 0.016 ms^-1

The recoil speed of the gun is 0.016 m s^-1.

Let first car be A and second be B,

Speed of car A, v_A = 18 km/h

Speed of car B, v_B = 27 km/h

Speed of bird, v = 36 km/h

Relative speed of cars,

v_AB = v_BA = v_A + v_B = 45 km/h

Initial distance between cars, x = 36 km

Time taken by cars to meet, t = \(\frac{x}{v_{AB}}=\frac{36}{45}\) km/h

Distance covered by bird in the time,

x' = v x t = 36 × 0.8 = 28.8 km

Displacement of bird = distance covered by car A in 0.8 h on straight line (because bird starts from car A)

∴ x_A = v_A x t = 18 × 0.8 = 14.4 km

Suppose man is at building A and wants to land on building B.

Vertical speed of man, v_y = 0

Relative speed of cars = 45km/h, time required to meet

= 36 km/45km/h = 0.80h

Thus, distance covered by the bird = 36 km/h × 0.8h = 28.8 km.

When the bus stops suddenly, people fall forward because their inertia as they are in state of motion even when the bus has come to rest.

(a) Mass

(b) forward

(c) backward

(d) inertia

(e) friction; air

p = m x v

This equation signifies that momentum of a body is the product of its mass and its velocity.

Here, p is momentum of the body

m is the mass of the body

v is the velocity of the body.

It is easier to stop a tennis ball than a cricket ball moving with the same speed because the momentum of cricket ball will be higher than the tennis ball as the mass of the cricket ball is more than tennis ball. Therefore, less force is required to stop the tennis ball when compared to the cricket ball.

Since the speed of tennis ball and cricket ball is same, the momentum of cricket ball will be higher due to its mass being greater than mass of tennis ball. So, less force is required to stop a tennis ball than to stop a cricket ball.

Being heavier, cricket ball has higher rate of change of momentum during motion so more force sumed.

This is because the mass of tennis ball is less than the cricket ball. So it has lesser momentum and hence smaller force is required to stop the tennis ball.

In transportation, the vehicle say a truck, may need to halt suddenly. To bring a fragile material, like porcelain object to a sudden halt means applying a large force and this is likely to damage the object. If it is wrapped up in say, straw, the object can travel some distance as the straw is soft before coming to a halt. The force needed to achieve this is less, thus reducing the possibility of damage.

So that inertia of motion may help him in his muscular efforts to take a longer jump.

(a) If both the assertion and the reason are true and the reason is the correct explanation of assertion.

This is because sand yields under the weight of the man. This difficulty can be overcome by placing wooden plank on the sand. Then the weight of the man will act on larger area and hence pressure exerted on sand decreases and the sand does not yield.