Categories: Jamia school entrance

JMI 11th Science Entrance Paper 2025 Answer Key

31. A piece of wire of resistance R is cut into five parts. These parts are then connected in parallel. If the equivalent resistance of this parallel combination is R’ then the ratio of R/R’ is

(A) R/R’ = 1/25
(B) R/R’ = 1/5
(C) R/R’ = 5
(D) R/R’ = 25

(D) R/R′=25

Explanation:

Resistance of each part is R/5. In parallel, 1/R′=5/(R/5)=25/R, so R′=R/25. Therefore, the ratio R/R′=25.

32. An electric bulb is rated 220V and 100W. When it is operated on 110V the power consumed will be

(A) P = 25W
(B) P = 5OW
(C) P = 15W
(D) P = 20W

(A) P=25W

Explanation:

Resistance R=V2/P=(220)2/100=484Ω. New power P′=(V′)2/R=(110)2/484=25 W.

33. Calculate the number of electrons constituting one coulomb of charge

(A) 6.50×10¹⁸
(C) 6.25×10¹⁸
(B) 6.00×10¹⁸
(D) 6.52×10¹⁸

Explanation:

Number of electrons n=Q/e=1C/(1.6×10−19C)=6.25×1018.

34. How much energy is given to each coulomb of charge passing through a 6V battery?

(A) E = 6J
(B) E = 12J
(C) E =18J
(D) E = 3J

(A) E=6J

Explanation:

Energy E=V×Q. For one coulomb (Q=1C) through 6V, E=6V×1C=6J.

35. An electron beam enters a magnetic field at right angles to it. In which magnetic field is going horizontally right ward and electron beam is perpendicular downward. The direction of force acting on the electron beam Will be.

Magnetic field —————–>‫‫
Electron beam ↓↓↓↓↓↓↓↓↓↓↓

(A) to the left
(B) to the right
(C) into the page
(D) out of the page

(D) out of the page

Explanation:

Using Fleming’s Left-Hand Rule, conventional current is opposite to electron flow (to the left). With the field into the page, the force is directed out of the page.

Page: 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20