The binomial expansion is given by the formula, 1 + x n = 1 + n x + n n - 1 x 2 2 ! + n n - 1 n - 2 x 3 3 ! + … Here, n = - 2 , therefore, the expansion will become as, 1 + x - 2 = 1 + - 2 x + - 2 - 2 - 1 x 2 2 ! + - 2 - 2 - 1 - 2 - 2 x 3 3 ! + … 1 + x - 2 = 1 - 2 x + 3 x 2 - 4 x 3 + … will be the expansion of 1 + x - 2 .

Using binomial expansion, simplify the expression Q 1 + ∆ x x 3 - 1 , assuming ∆ x to be small in comparison to x .

The given expression is Q 1 + ∆ x x 3 - 1 By the binomial expansion, 1 + y n = 1 + n y + n n - 1 y 2 2 ! + n n - 1 n - 2 y 3 3 ! + … So, 1 + ∆ x x 3 = 1 + 3 ∆ x x + 3 3 - 1 2 ! ∆ x x 2 + . . . . Neglecting higher-order terms, we get, 1 + ∆ x x 3 ≈ 1 + 3 ∆ x x . . . 1 Putting the value of equation 1 in the initial expression Q 1 + ∆ x x 3 - 1 = Q 1 + 3 ∆ x x - 1 = 3 Q ∆ x x .

If x is small enough, show that 1 + x 1 - x ≈ 1 + 2 x .

Here given expression is 1 + x 1 - x ⇒ 1 + x 1 - x = 1 + x 1 - x - 1 … 1 By the binomial expansion, 1 + y n = 1 + n y + n n - 1 y 2 2 ! + n n - 1 n - 2 y 3 3 ! + ⋯ … 2 Using equation 2 , we get, 1 - x - 1 = 1 + - 1 - x + - 1 - 1 - 1 - x 2 + ⋯ ⇒ 1 - x - 1 = 1 + x … 3 By equation 1 and 3 , we get, 1 + x 1 - x = 1 + x 1 + x = 1 + 2 x + x 2 ∵ x is small so, x 2 = 0 ⇒ 1 + x 1 - x ≈ 1 + 2 x

The mass of m of a body moving with a velocity v is given by m = m 0 1 - v 2 c 2 , where m 0 = rest mass of body = 10 kg and c = speed of light = 3 × 10 8 m s - 1 . Find the value of m at v = 3 × 10 7 m s - 1 .

The relativistic mass of the particle is m = m 0 1 - v 2 c 2 . ⇒ m = m 0 1 - v 2 c 2 - 1 2 … 1 By the binomial expansion, 1 + y n = 1 + n y + n n - 1 y 2 2 + n n - 1 n - 2 y 3 3 + ⋯ Now, let us apply the expansion formula for 1 - v 2 c 2 - 1 2 Here y = - v c 2 ; n = - 1 2 Now, putting these values in the binomial expansion formula, 1 - v 2 c 2 - 1 2 = 1 + - 1 2 - v 2 c 2 + ⋯ Neglecting higher-order terms, 1 - v 2 c 2 - 1 2 ≈ 1 + v 2 2 c 2 … 2 ∵ v = 3 × 10 7 m s - 1 and c = 3 × 10 8 m s - 1 Putting these values in equation 2 , we get, 1 - v 2 c 2 - 1 2 ≈ 1 + 3 × 10 7 2 2 3 × 10 8 2 ⇒ 1 - v 2 c 2 - 1 2 ≈ 1 + 1 200 ≈ 1 + . 005 ≈ 1 . 005 … 3 By equation 3 and 1 , m = m 0 1 . 005 ∵ Rest mass of the particle = m 0 = 10 kg So, m = 10 × 1 . 005 = 10 . 05 kg Hence, the value of the mass is 10 . 05 kg .

E M B I B E

PHYSICS For Joint Entrance Examination JEE (Advanced) Mechanics I>Chapter 2 - Basic Mathematics>CONCEPT APPLICATION EXERCISE 2.1

B M Sharma Solutions for Chapter: Basic Mathematics, Exercise 4: CONCEPT APPLICATION EXERCISE 2.1

Author:B M Sharma

B M Sharma Physics Solutions for Exercise - B M Sharma Solutions for Chapter: Basic Mathematics, Exercise 4: CONCEPT APPLICATION EXERCISE 2.1

Attempt the practice questions on Chapter 2: Basic Mathematics, Exercise 4: CONCEPT APPLICATION EXERCISE 2.1 with hints and solutions to strengthen your understanding. PHYSICS For Joint Entrance Examination JEE (Advanced) Mechanics I solutions are prepared by Experienced Embibe Experts.

Questions from B M Sharma Solutions for Chapter: Basic Mathematics, Exercise 4: CONCEPT APPLICATION EXERCISE 2.1 with Hints & Solutions

EASY

JEE Advanced

IMPORTANT

PHYSICS For Joint Entrance Examination JEE (Advanced) Mechanics I>Chapter 2 - Basic Mathematics>CONCEPT APPLICATION EXERCISE 2.1>Q 1

Expand ${(1 + x)}^{- 2}$ .

MEDIUM

JEE Advanced

IMPORTANT

PHYSICS For Joint Entrance Examination JEE (Advanced) Mechanics I>Chapter 2 - Basic Mathematics>CONCEPT APPLICATION EXERCISE 2.1>Q 2

Using binomial expansion, simplify the expression $Q [{(1 + \frac{∆ x}{x})}^{3} - 1]$ , assuming $∆ x$ to be small in comparison to $x$ .

EASY

JEE Advanced

IMPORTANT

PHYSICS For Joint Entrance Examination JEE (Advanced) Mechanics I>Chapter 2 - Basic Mathematics>CONCEPT APPLICATION EXERCISE 2.1>Q 3

Write down the first four terms of the expansion ${(1 + x^{2})}^{- 2}$ .

EASY

JEE Advanced

IMPORTANT

PHYSICS For Joint Entrance Examination JEE (Advanced) Mechanics I>Chapter 2 - Basic Mathematics>CONCEPT APPLICATION EXERCISE 2.1>Q 4

If $x$ is small enough, show that $\frac{1 + x}{1 - x} \approx 1 + 2 x$ .

EASY

JEE Advanced

IMPORTANT

PHYSICS For Joint Entrance Examination JEE (Advanced) Mechanics I>Chapter 2 - Basic Mathematics>CONCEPT APPLICATION EXERCISE 2.1>Q 5

The mass of $m$ of a body moving with a velocity $v$ is given by $m = \frac{m_{0}}{\sqrt{1 - \frac{v^{2}}{c^{2}}}}$ , where $m_{0} =$ rest mass of body $= 10 kg$ and $c =$ speed of light $= 3 \times 10^{8} m s^{- 1}$ . Find the value of $m$ at $v = 3 \times 10^{7} m s^{- 1}$ .

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