Correct Answer: A. I and III only

Explanation

In SHM, the body must have acceleration in a direction opposite to the displacement and the acceleration must be directly proportional to displacement. Also, the acceleration must be directed towards its mean position. This is not the case with a football on the pitch.

Correct Answer: B. National Congress of British West Africa

Explanation

The National Congress of British West Africa (NCBWA), founded in 1917, was one of the earliest nationalist organizations in West Africa, and one of the earliest formal organizations working toward African emancipation.

Correct Answer: (c) tremulous

Correct Answer: A. encourages entrepreneuership

Correct Answer: B. 6hr 15min

Explanation

Total capacity of the container = 45 liters = 45 x 1000

= 45000 cm\(^3\)

Time to fill the container = \(\frac{45000}{2}\)

= 22500 seconds

= \(\frac{22500}{3600}\)

= 6 hr 15 mins

Correct Answer: D) pipes

Explanation

F = \(kv^xρ^yA^z\)

For the left-hand side:

F = mass × acceleration = \(MLT^{-2}\)

For the right-hand side:

v = \(LT^{-1}\), ρ = \(ML^{-3}\) and A = \(L^2\)

So,

\(MLT^{-2} = k(LT^{-1})^x(ML^{-3})^y(L^2)^z\)

Since k is dimensionless,

\(MLT^{-2} = (LT^{-1})^x(ML^{-3})^y(L^2)^z\)

\(MLT^{-2} = L^x × T^{(-1)x} \times M^y × L^{(-3)y} \times L^{(2)z}\)

\(MLT^{-2} = L^x × T^{-x} \times M^y \times L^{-3y} \times L^{2z}\)

\(MLT^{-2} = L^{(x - 3y + 2z)} \times T^{-x} \times M^y\)

\(M^1L^1T^{-2}= L^{(x - 3y + 2z)} \times T^{-x} \times M^{y}\)

Comparing the powers:

For M,

y = 1

For L,

x - 3y + 2z = 1 ---- (i)

For T,

-x = -2

∴ x = 2

Substitute (2) for x and (1) for y in equation (i)

⇒ 2 - 3(1) + 2z = 1

⇒ 2 - 3 + 2z = 1

⇒ -1 + 2z = 1

⇒ 2z = 1 + 1

⇒ 2z = 2

∴ z = 1

Hence, x = 2, y = 1 and z = 1.

Correct Answer: C. \(\frac{79}{156}\)

Explanation

Sin x = \(\frac{5}{13}\)

0^o \(\leq\) x \(\leq\) 90^o, (cos x - tan x)

AC² = AB² + BC²

13² = 5² + BC²

169 - 25 + BC²

169 - 25 = BC²

144 = BC²

Cos x = \(\frac{Adj}{Hyp}\) = \(\frac{12}{13}\)

BC = \(\sqrt{144}\)

BC = 12

tan x = \(\frac{opp}{adj} = \frac{5}{12}\)

BC = 12

cos x - tan x = \(\frac{12}{13} - \frac{5}{12}\)

\(\frac{144 - 65}{156} = \frac{79}{156}\)

Correct Answer: D. scales

Explanation

\(h = 20t - \frac{3}{2}t^{2} - \frac{2}{3}t^{3}\)

Velocity, \(\frac{\mathrm d h}{\mathrm d t} = 20 - 3t - 2t^{2}\)

At point of rest, v = 0

\(2t^{2} + 3t - 20 = 0\)

\(\implies 2t^{2} - 5t + 8t - 20 = 0\)

\(t(2t - 5) + 4(2t - 5) = 0 \implies t = \frac{5}{2}; -4\)

The value of time cannot be negative hence \(t = \frac{5}{2}secs\)

(b) Maximum height, \(h_{m}\) is at time \(t = \frac{5}{2}s\).

\(h_{m} = 20(\frac{5}{2}) - \frac{3}{2}(\frac{5}{2})^{2} - \frac{2}{3}(\frac{5}{2})^{3}\)

= \(50 - \frac{75}{8} - \frac{250}{16} = \frac{400}{16} = 25 m\)