JEE Advanced (Matrix Match & Integer Answer): Units & Measurements | Chapter-wise Tests for JEE Main & Advanced PDF Download

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Match the Following

DIRECTIONS (Q. No. 1) : Each question contains statements given in two columns, which have to be matched. The statements in Column-I are labelled A, B, C and D, while the statements in Column-II are labelled p, q, r and s. Any given statement in Column-I can have correct matching with ONE OR MORE statement(s) in Column-II. The appropriate bubbles corresponding to the answers to these questions have to be darkened as illustrated in the following example :

If the correct matches are A-p, s and t; B-q and r; C-p and q; and D-s then the correct darkening of bubbles will look like the given.

Q.1. Some physical quantities are given in Column I and some possible SI units in which these quantities may be expressed are given in Column II. Match the physical quantities in Column I with the units in Column II and indicate your answer by darkening appropriate bubbles in the 4 × 4 matrix given in the ORS.      (2007)

Column I	Column II
(A) GMeMs , G – universal gravitational constant,   Me – mass of the earth, Ms – mass of the Sun	(p) (volt) (coulomb)(metre)
(B) 3RT/M ,  R – universal gas constant,  T – absolute temperature, M – molar mass	(q) (kilogram) (metre)3(second)–2
F – Force, q – charge, B – magnetic field	(r) (metre)2 (second)–2
Me – mass of the earth, Re – radius of the earth	(s) (farad) (volt)2 (kg)–1

Ans. (A) → p, q ; (B) → r, s ;  (C) → r, s ; (D) → r,s

Solution. A : p → q

Reason : Unit of GM_eM_s = Fr² = Nm²

= kg m³s^–2
Also (volt) (coulomb) (metre) = (joule) (metre)
= (N - m) (m) = Nm² = kg m³s^–2

B : r → s

C : r → s

∴ Unit of v² is m²s^–2 which is further equal to FV² kg^–1.

D : r → s

DIRECTIONS (Q. No. 2) : Following question has matching lists. The codes for the lists have choices (a), (b), (c) and (d) out of which ONLY ONE is correct.

Q.2. Match List I with List II and select the correct answer using the codes given below the lists:       (JEE Adv. 2013)

Codes:

Ans. (c)

Solution. 

(c) is the correct option.

Integer Value Correct Type 

Q.1. To find the distance d over which a signal can be seen clearly in foggy conditions, a railways-engineer uses dimensions and assumes that the distance depends on the mass density ρ of the fog, intensity (power/area) S of the light from the signal and its frequency f. The engineer finds that d is proportional to S^1/n. The value of n is       (JEE Adv. 2014)

Ans. (3)

Solution. 

Q.2. During Searle’s experiment, zero of the Vernier scale lies between 3.20 × 10^–2 m and 3.25 × 10^–2 m of the main scale. The 20th division of the Vernier scale exactly coincides with one of the main scale divisions. When an additional load of 2 kg is applied to the wire, the zero of the Vernier scale still lies between 3.20 × 10^–2 m and 3.25 × 10^–2 m of the main scale but now the 45^th division of Vernier scale coincides with one of the main scale divisions. The length of the thin metallic wire is 2 m and its cross-sectional area is 8 × 10^–7 m². The least count of the Vernier scale is 1.0 × 10^–5 m. The maximum percentage error in the Young’s modulus of the wire is        (JEE Adv. 2014)

Ans. (4)

Solution. 

Here F, a and L are accurately known.

Q.3. The energy of a system as a function of time t is given as E(t) = A² exp(–αt,) where a = 0.2 s^–1. The measurement of A has an error of 1.25%. If the error in the measurement of time is 1.50%, the percentage error in the value of E(t) at t = 5 s is     (JEE Adv. 2015)

Ans. (4)

Solution. E = A² e–0.2t
∴ log_e E = 2 loge A –0.2t
On differentiating we get

As errors always add up therefore