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Accounting equation is A = C + L

∴ Cash + Other Assets = Capital + Liabilities

∴ Cash + 60,000 = 40,000 + 50,000

∴ Cash + 60,000 = 90,000

∴ Cash = 90,000-60,000 = 30,000

∴ Amount of cash is ₹ 30,000.

Effects :

(i) Bank is receiver → Bank balance (Asset) increases
(ii) Dividend is an income → Capital increases

Conclusion :

→ Assets and Capital have increased by the same amount of ₹ 1,200.
→ Thus, equation is satisfied.

Effects :

(i) Bank is receiver → Bank balance (Asset) increases
(ii) Cash goes out → Cash (Asset) decreases

Conclusion :

→ Assets have increased and decreased by the same amount of ₹ 10,000.
→ Thus, equation is satisfied.

Given c = 4π, \(\frac{dc}{dt}\)= 3cm/sec \(\frac{dA}{dt}\)= ? \(\frac{dr}{dt}\)= ?

Circumference = c = 2πr 

4π = 2πr ⇒ r = 2 

Again 

C = 2πr & A = πr² 

\(\frac{dc}{dt}\) = 2π \(\frac{dr}{dt}\).\(\frac{dA}{dt}\) = π . 2r.\(\frac{dr}{dt}\) 

3 = 2π . \(\frac{dr}{dt}\) = π . 2. 2. \(\frac{3}{2\pi}\)

⇒ \(\frac{dr}{dt}\) = \(\frac{3}{2\pi}\)cm/ sec \(\frac{dA}{dt}\) = 6cm²/sec

Blotting paper has fine pores which act as capillaries. The ink rises in these capillaries. Thus, the ink is absorbed by the blotting paper.

Correct option is D) 0

Correct option is D) 0

The number of electrons present in the outermost orbit of an atom is called its valency.

Correct option is A) 2, 3

Correct option is A) 8

Correct option is C) M

Correct option is B) L

Lagrange’s mean value theorem states that if a function f(x) is continuous on a closed interval [a, b] and differentiable on the open interval (a, b), then there is at least one point x = c on this interval, such that

f(b) − f(a) = f′(c)(b − a)

\(\Rightarrow f'(c)=\frac{f(b) - f(a)}{b-a}\)

This theorem is also known as First Mean Value Theorem.

f(x) = 2x – x² on [0, 1]

Every polynomial function is continuous everywhere on (−∞, ∞) and differentiable for all arguments.

Here, f(x) is a polynomial function. So it is continuous in [0, 1] and differentiable in (0, 1). So both the necessary conditions of Lagrange’s mean value theorem is satisfied.

Therefore, there exists a point c∈(0, 1) such that:

\(f'(c)=\frac{f(1) - f(0)}{1-0}\)

⇒ f’(c) = f(1) – f(0)

f(x) = 2x – x²

Differentiating with respect to x:

f’(x) = 2 – 2x

For f’(c), put the value of x = c in f’(x):

f’(c)= 2 – 2c

For f(1), put the value of x = 1 in f(x):

f(1) = 2(1) – (1)²

= 2 – 1

= 1

For f(0), put the value of x = 0 in f(x):

f(0) = 2(0) – (0)²

= 0 – 0

= 0

f’(c) = f(1) – f(0)

⇒ 2 – 2c = 1 – 0

⇒ – 2c = 1 – 2

⇒ – 2c = – 1

\(\Rightarrow c=\frac{-1}{-2}=\frac{1}{2}∈(0, 1)\)

Hence, Lagrange’s mean value theorem is verified.

Payload of the balloon = mass of the displaced air - mass of the balloon.

Radius of the balloon, r = 10 m

Mass of the balloon, m = 100 kg

Volume of the balloon

= 4/3 πr³

= 4/3 π x (10 m)³

Therefore, Volume of the displaced air = 4188.8 m³

and mass of the displaced air = 4188.8 m³ x 1.2 kg m^-3 = 5026.6 kg

Let W be the mass of helium gas filled into the balloon, then

pV = (W/M)RT

or, W = {pVM}/{RT}

= {1.66 bar x 4188.8 x 10³ dm³ x 4 g mol^-1}/{0.083 bar dm³ K^-1 mol^-1 x 300 K}

or, W = 1.117 x 10⁶ g

= 1117.0 kg

Total mass of the balloon filled with Helium

= 1117.0 kg + 100 kg

= 1217.0 kg

Then, payload of the balloon

= 5026.6 kg - 1217.0 kg

= 3809.6 kg

The correct option is C) γ-rays.

γ has the highest penetrating power whereas alpha ray has the least penetrating power.

1. The distance between P and O (the school) is less then the distance between Q and O. (from the graph)

⇒ A lives closer to the school than B.

2. From the graph, we see that the line for A starts before the line for B.

⇒ A starts from the school earlier than B.

3. Velocity = Slope of the x-t graph. Clearly, the slope for B > slope for A.

⇒ B walks faster than A

4. From the graph, we see that both lines end at the same time.

⇒ A and B reach home at the same time.

5. From the graph, we see that the lines intersect once. Also, B walks faster.

⇒ B overtakes A on the road once.

Correct option is a. ¹⁴₆C, ¹⁵₇N, ¹⁷₉F

(2) an Arithmetic Progression

If t₁, t₂, t₃, … is 1, 2, 3, …

If t₆ = 6, t₁₂ = 12, t₁₈ = 18 then 6, 12, 18 … is an arithmetic progression

A’s speed = 15 km/h

B’s speed = 12 km/h

Distance covered by A in 6 hours = 15 x 6 = 90 km

and Distance covered by B in 6 hours = 12 x 6 = 72 km

(i) Distance between A and B when they move in the same direction = 90 – 72 = 18 km

(ii) Distance between A and B, when they move in the opposite directions = 90 + 72 = 162 km

Let the number of tickets sold of Rs 100 be x. 

The number of tickets of Rs 200 sold was 20 more than the number of tickets of Rs 100. 

∴ Number of tickets sold of Rs 200 = (x + 20) 

∴ Total amount received by the theatre through the sale of tickets 

= 100 × x + 200 × (x + 20) = 100x + 200x + 4000 = 300x + 4000 

Since, the total amount received by the theatre through the sale of tickets = Rs 37000 

∴ 300x + 4000 = 37000 

∴ 300x = 37000 – 4000

∴ 300x = 33000

∴ x = 110 

∴ 110 tickets of Rs 100 were sold.

Let the digit at unit’s place be x. 

The digit at the ten’s place is twice the digit at unit’s place. 

∴ The digit at ten’s place = 2x

Since, the sum of the original number and the new number is 66. 

∴ 21x + 12x = 66 

∴ 33x = 66

∴ x = 2 

∴ Original number = 21x = 21 × 2 = 42 

∴ the original number is 42.

The ‘Internet’ or the net is a network of computers.

Looking at various websites in computers is known as ‘browsing’ or surfing.

• CD: Computer Disks 
• ROM: Read Only Memory 
• RAM: Random Access Memory 
• PPT: Power Point 
• e-mail: Electronic mail 
• IP: Internet Protocol 
• EC: Electronic Commerce 
• EDI: Electronic Data Interchange 
• E-Market: Electronic Market 
• VAN: Value Added Network 
• E-map: Electronic map

Yes, I would like to be a netizen and explore this ocean of information.

People who use the net are ‘netizens’.

A ‘password’ is a secret word or group of letters and/ or numbers which will allow a person to use or see the email account on typing it.

Being ‘offline’ means we are not connected to the net.

(a) as e-mail or electronic mail. 

(b) e-mail addresses

1. steady × unsteady 

2. known × unknown 

3. expected × unexpected 

4. firm × shaky

1. 1. unknown 

2. unlucky

2. 1. Satyendra Nath Bose was one of the known scientists of India. 

2. Our school team was lucky to win the Kabaddi match at the last moment.

Indian railway runs on multiple gauge operations, i.e. broad gauge, metre gauge and narrow gauge. Broad gauge and narrow gauge can be differentiated based on their measurements and the distance they can be used to travel. Broad gauge is 1.676 metres, and the narrow gauge is 0.762 and 0.610. Running track in broad gauge is 77,347 km, and in narrow gauge, it is 2474 km.

(a) Project Unigauge

(b) Inland Waterways Authority of India (IWAI)

We know that two or more lines are said to be concurrent if they intersect at a single point.

We first find the point of intersection of the first two lines.

2x + 5y = 1 ....(1)

x - 3y = 6 ....(2)

Multiplying (2) by 2, we get,

2x − 6y = 12 ....(3)

Subtracting (3) from (1), we get,

11y = −11

y = −1

From (2), x = 6 + 3y = 6 − 3 = 3

So, the point of intersection of the first two lines is (3, −1).

If this point lie on the third line, i.e., x + 5y + 2 = 0, then the given lines will be concurrent. Substituting x = 3 and y = -1, we have:

L.H.S. = x + 5y + 2

= 3 + 5(-1) + 2

= 5 - 5

= 0 = R.H.S.

Thus, (3, −1) also lie on the third line.

Hence, the given lines are concurrent.

Ethyl Alcohol.

Given equation of a line is x - y = 4

⟹ y = x - 4

Comparing this equation with y = mx + c. We have:

Slope = m = 1

y-intercept = c = −4

Let the inclination be θ.

Slope = 1 = tan θ = tan 45^o

∴ θ = 45°

Given, y-intercept = c = −1 and inclination = 45^o.

Slope = m = tan 45^o = 1

Substituting the values of c and m in the equation y = mx + c, we get,

y = x - 1, which is the required equation.

Given m = 20000 kg

a = 5ms^-2 (against gravity) since the rocket has to move upwards against gravity the total initial thrust of the blast is given by

F = ma + mg

= m (a + g) = 20000 (5 + 9.8)

= 296 × 10⁵N.

Mass of the rocket, m = 20,000 kg
Initial acceleration, a = 5 m/s²
Acceleration due to gravity, g = 10 m/s²
Using Newton’s second law of motion, the net force (thrust) acting on the rocket is given by the relation:
F – mg = ma
F = m (g + a)
= 20000 × (10 + 5) = 20000 × 15 = 3 x10⁵ N

(a) Vertically downward
Parabolic path
At the extreme position, the velocity of the bob becomes zero. If the string is cut at this moment, then the bob will fall vertically on the ground.
(b) At the mean position, the velocity of the bob is 1 m/s. The direction of this velocity is tangential to the arc formed by the oscillating bob. If the bob is cut at the mean position, then it will trace a projectile path having the horizontal component of velocity only.
Hence, it will follow a parabolic path.

(a) As the bob of simple pendulum has no velocity at, the extreme position, it will just fall down when the string is cut at extreme position.

(b) At mean position, if we cut the string, the bob will have periodic path as it is having horizontal velocity.

It will reach the bottom with same speed in each case because speed depends upon height and not in inclination.

It will take longer to roll down on the plane with smaller inclination.

1 : 2 is the ratio of their masses.

A torque is required to roll the given disc. As per the definition of torque, the rotating force should be tangential to the disc. Since the frictional force at point B is along the tangential force at point A, a frictional force is required for making the disc roll.

Force of friction acts opposite to the direction of velocity at point B. The direction of linear velocity at point B is tangentially leftward. Hence, frictional force will act tangentially rightward. The sense of frictional torque before the start of perfect rolling is perpendicular to the plane of the disc in the outward direction.

Since frictional force acts opposite to the direction of velocity at point B, perfect rolling will begin when the velocity at that point becomes equal to zero. This will make the frictional force acting on the disc zero.

The conventional rules are:

(i) If the insignificant digit to be dropped is more than 5, the preceeding digit is increased by 1, but if it is less than five, then preceeding digits is not changed, e.g., 1.748 is rounded off to 3 significant figures as 1.75 and 1.742 as 1.74.

(ii) If the insignificant digit to be dropped is 5, then this digit is simply dropped if the preceeding digit is increased by 1.

E.g., the number 1.845 rounded off to three significant digits is 1.84 but for number 1.875 it is 1.88.

Using, f = μmgcos θ,

We get \(f=\frac{1}{3}\times 10 \times 9.8 \times cos\, 30°\)

\(=\frac{1}{3}\times 10 \times 9.8 \times \frac{√3}{2}\)

= 28.25 N

Gravitational force and electrostatic force are the examples of conservative force.

Yes, when K < U, total energy E = K + U is negative. The body has the momentum as K ≠ 0.

for examples, in an atom, electron has momentum through its energy is negative.

Yes, when p = 0, K = P²/2m = 0

But, E = K + U = U (P.E) which may or may not be zero.

The momentum of the meteorite is transferred to the air molecules. Thus, the principle of momentum conservation is not violated.