JMI 11th Science Entrance Paper 2023 Answer Key

36. In the given figure, O is the centre of the circle. The area of the sector APB is 5/18 of the area of the circle. The value of x is:

(A) $60 °$

(B) $70 °$

(D) $100 °$

Answer:

(D)

100 °

Explanation:

Area of a sector is given by

\frac{x}{360}\pi r^2

. Setting this equal to

\frac{5}{18}\pi r^2

, the

\pi r^2

terms cancel out. Solving for x gives

\frac{5}{18}*360 = 100

37. The maximum volume of a cone that can be carved out of a solid hemisphere of radius “r” is:

(A) $3\pi r^2$

(B) $\frac{1}{3}\pi r^3$

(D) $3\pi r^3$

Answer:

\frac{1}{3}\pi r^3

Explanation:

The maximum cone inside a hemisphere shares the same base radius (r) and has a height equal to the hemisphere’s radius (r).
Cone volume is

\frac{1}{3}\pi r^2h = \frac{1}{3}\pi r^3

38. If $\alpha, \beta$ are the roots of the equation $x^2 -P(x+1) +C = 0$ , then $(\alpha + 1) (\beta +1) = ?$

(A) $C-1$

(B) $1-C$

(D) $1+ C$

Answer:

(D)

1+ C

Explanation:

Expanding the quadratic gives

x^2 -Px +(C-P) = 0

. The sum of roots

\alpha + \beta =P

, and product

\alpha * \beta =C -P

. Expanding

(\alpha +1) (\beta + 1)= \alpha \beta + \alpha + \beta +1

. Substituting yields

(C-P) + P+1 = C +1

39. If the numbers a, b, c, d, e form an AP, then the value of $a-4b+6c-4d + e$ is:

(A) 1

(B) 2

(D) None of these

Answer:

Explanation:

By defining the AP terms around the middle term as

a = c-2D, b=c-D, c=c, d=c+D, e=c+2D

, and substituting these into the expression, all ‘c’ and ‘D’ terms perfectly cancel out, evaluating to 0.

40. All squares are:

(A) Congruent

(B) Similar

(D) None of these

Answer:

Explanation:

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