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

 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 :

JEE Advanced (Matrix Match & Integer Answer): Current Electricity | Chapter-wise Tests for JEE Main & Advanced

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. Column I gives some devices and Column II gives some processes on which the functioning of these devices depend. Match the devices in Column I with the processes in Column II and indicate your answer by darkening appropriate bubbles in the 4 × 4 matrix given  in the  ORS.

Column I

Column II

(A) Bimetallic strip

(p) Radiation from a hot body

(B) Steam engine
(q) Energy conversion
(C) Incandescent lamp

(r) Melting

(D) Electric fuse
(s) Thermal expansion of solids

Ans. A → s; B → q; C → p, q; D →q, r

Solution. 

A → s 

Reason : Bimetallic strip is based on thermal expansion of solids.

B → q 

Steam engine is based on energy conversion.

C → p, q 

Incandescent lamp is based on energy conversion and radiation from a hot body.

D → q, r 

Electric fuse is based on melting point of the fuse material which is turn depends on the heating effect of current.

Q.2. When two identical batteries of internal resistance 1Ω each are connected in series across a resistor R, the rate of heat produced in R is J1. When the same batteries are connected in parallel across R, the rate is J2. If J1 = 2.25 J2 then the value of R in Ω is

Ans.  4

Solution. Cells connected in series

JEE Advanced (Matrix Match & Integer Answer): Current Electricity | Chapter-wise Tests for JEE Main & Advanced

JEE Advanced (Matrix Match & Integer Answer): Current Electricity | Chapter-wise Tests for JEE Main & Advanced

JEE Advanced (Matrix Match & Integer Answer): Current Electricity | Chapter-wise Tests for JEE Main & Advanced

JEE Advanced (Matrix Match & Integer Answer): Current Electricity | Chapter-wise Tests for JEE Main & Advanced

JEE Advanced (Matrix Match & Integer Answer): Current Electricity | Chapter-wise Tests for JEE Main & Advanced

Q.3. At time t = 0, a battery of 10 V is connected across points A and B in the given circuit. If the capacitors have no charge initially, at what time (in sceonds) does the voltage across them become 4 V? [Take : ln5 =1.6, ln3 = 1.1]

JEE Advanced (Matrix Match & Integer Answer): Current Electricity | Chapter-wise Tests for JEE Main & Advanced

Ans. 2

Solution. The equivalent circuit is shown in the figure.

JEE Advanced (Matrix Match & Integer Answer): Current Electricity | Chapter-wise Tests for JEE Main & Advanced

∴The time constant τ = RC = 4 sec The potential across 4μF capacitor at any time ‘t’ is given as

JEE Advanced (Matrix Match & Integer Answer): Current Electricity | Chapter-wise Tests for JEE Main & Advanced

Q.4. Two batteries of different emfs and different internal resistances are connected as shown. The voltage across AB in volts is

JEE Advanced (Matrix Match & Integer Answer): Current Electricity | Chapter-wise Tests for JEE Main & Advanced

Ans. 5

Solution. Let i be the current flowing in the circuit. Apply Kirchhoff’s law in the loop we get

–3 – 2i – i + 6 = 0

∴ 3i = 3 

∴ i = 1 Amp

Now let us travel in the circuit from A to B through battery of 6V, we get 

VA – 6 + 1 × 1 = VB

∴ V– VB = 5 volt.

Q.5.A galvanometer gives full scale deflection with 0.006 A current. By connecting it to a 4990 W resistance, it can be converted into a voltmeter of range 0 – 30 V. If connected to 2n/249Ω resistance, it becomes an ammeter of range 0 – 1.5A.The value of n is

Ans. 5

Solution.

 JEE Advanced (Matrix Match & Integer Answer): Current Electricity | Chapter-wise Tests for JEE Main & Advanced

∴n ≈ 5

Q.6. In the following circuit, the current through the resistor R (= 2 Ω) is I amperes. The value of I is

JEE Advanced (Matrix Match & Integer Answer): Current Electricity | Chapter-wise Tests for JEE Main & Advanced

Ans. 1

Solution. The equivalent resistance of balanced wheatstone bridge is

JEE Advanced (Matrix Match & Integer Answer): Current Electricity | Chapter-wise Tests for JEE Main & Advanced

The equivalent resistance of balanced wheat stone bridge is

JEE Advanced (Matrix Match & Integer Answer): Current Electricity | Chapter-wise Tests for JEE Main & Advanced

JEE Advanced (Matrix Match & Integer Answer): Current Electricity | Chapter-wise Tests for JEE Main & Advanced

The document JEE Advanced (Matrix Match & Integer Answer): Current Electricity | Chapter-wise Tests for JEE Main & Advanced is a part of the JEE Course Chapter-wise Tests for JEE Main & Advanced.
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