JEE Advanced Mix Test 35

Result

JEE Advanced Mix Test 35

Score
-
out of -
Rank
-
out of -

TIME Taken - -

SHARING IS CARING

If our Website helped you a little, then kindly spread our voice using Social Networks. Spread our word to your readers, friends, teachers, students & all those close ones who deserve to know what you know now.

Weekly Quiz Competition

Question 1
4 / -1

In front of an earthed conductor a point charge + q is placed as shown in figure:

A
On the surface of conductor the net charge is always negative.

B
On the surface of conductor at some points charges are negative and at some points charges may be positive distributed non uniformly

C
Inside the conductor electric field due to point charge is non zero

D
None of these

Solution

Charge is distributed over the surface of conductor in such a way that net field due to this charge and outside charge q is zero inside the conductor. Field due to only q is non-zero.
Question 2
4 / -1

In the figure shown the key is switched on at t = 0. Let I₁ and I₂ be the currents through inductors having self inductances L₁ & L₂ at any time t respectively. The magnetic energy stored in the inductors 1 and 2 be U₁ and U₂. Then at any instant of time is :

A

B

C

D

Solution
Question 3
4 / -1

Statement – 1 Four point charges q₁, q₂, q₃ and q₄ are as shown in figure. The flux over the shown Gaussian surface depends only on charges q₁ and q₂.

Statement – 2 Electric field at all points on Gaussian surface depends only on charges q₁ and q₂.

A
Statement-1 is True, Statement-2 is True; Statement-2 is a correct explanation for Statement-1

B
Statement-1 is True, Statement-2 is True; Statement-2 is NOT a correct explanation for Statement-1

C
Statement-1 is True, Statement-2 is False

D
Statement-1 is False, Statement-2 is True.

Solution

Statement ? is true directly from Gauss Theorem.

Statement 2 is false. Electric field at any point an Gaussian surface depends on all four charges.

Statement-1 is True, Statement-2 is False
Question 4
4 / -1

Statement 1 : A direct uniformly distributed current flows through a solid long metallic cylinder along its length. It produces magnetic field only outside the cylinder.

Statement 2 : A thin long cylindrical tube carrying uniformly distributed current along its length does not produce a magnetic field inside it. Moreover, a solid cylinder can be supposed to be made up of many thin cylindrical tubes.

A
Statement-1 is True, Statement-2 is True; Statement-2 is a correct explanation for Statement-1.

B
Statement-1 is True, Statement-2 is True; Statement-2 is NOT a correct explanation for Statement-1

C
Statement-1 is True, Statement-2 is False

D
Statement-1 is False, Statement-2 is True

Solution

The current through solid metallic cylinder also produces magnetic field inside the cylinder. Hence statement-1 is false
Question 5
4 / -1

Statement–1 : Magnitude of potential difference across the terminals of a non-ideal battery in a circuit cannot be greater than its emf.

Statement–2 : When a current of magnitude I is passing through a battery of emf E and internal resistance r as shown, the magnitude of potential difference (V) across the battery is given by V = E– I r

A
Statement-1 is True, Statement-2 is True; Statement-2 is a correct explanation for Statement-1

B
Statement-1 is True, Statement-2 is True; Statement-2 is NOT a correct explanation for Statement-1

C
Statement-1 is True, Statement-2 is False

D
Statement-1 is False, Statement-2 is True.

Solution

Statement-1 is obviously false if the current is sent in opposite direction given in figure of statement-2
Question 6
4 / -1

A
Statement-1 is True, Statement-2 is True; Statement-2 is a correct explanation for Statement-1

B
Statement-1 is True, Statement-2 is True; Statement-2 is NOT a correct explanation for Statement-1

C
Statement-1 is True, Statement-2 is False

D
Statement-1 is False, Statement-2 is True.

Solution

along any closed path within a uniform magnetic field is always zero. Hence the closed path can be chosen of any size, even very small size enclosing a very small area. Hence we can prove that net current through each area of infinitesimally small size within region of uniform magnetic field is zero. Hence we can say no current (rather than no net current) flows through region of uniform magnetic field. Hence statement -2 is correct explanation of statement-1.
Question 7
4 / -1

In the circuit given below, both batteries are ideal. EMF E₁ of battery 1 has a fixed value, but emf E₂of battery 2 can be varied between 1.0 V and 10.0 V. The graph gives the currents through the two batteries as a function of E₂, but are not marked as which plot corresponds to which battery. But for both plots, current is assumed to be negative when the direction of the current through the battery is opposite the direction of that battery's emf. (Direction of emf is from negative to positive)

The value of emf E1 is :

A
8 V

B
6 V

C
4 V

D
2 V

Solution
Question 8
4 / -1

In the circuit given below, both batteries are ideal. EMF E₁ of battery 1 has a fixed value, but emf E₂ of battery 2 can be varied between 1.0 V and 10.0 V. The graph gives the currents through the two batteries as a function of E₂, but are not marked as which plot corresponds to which battery. But for both plots, current is assumed to be negative when the direction of the current through the battery is opposite the direction of that battery's emf. (Direction of emf is from negative to positive)

The resistance R₁ has value :

A
10 Ω

B
20 Ω

C
30 Ω

D
40 Ω

Solution
Question 9
4 / -1

In the circuit given below, both batteries are ideal. EMF E₁ of battery 1 has a fixed value, but emf E₂of battery 2 can be varied between 1.0 V and 10.0 V. The graph gives the currents through the two batteries as a function of E₂, but are not marked as which plot corresponds to which battery. But for both plots, current is assumed to be negative when the direction of the current through the battery is opposite the direction of that battery's emf. (Direction of emf is from negative to positive)

The resistance R₂ is equal to :

A
10 Ω

B
20 Ω

C
30 Ω

D
40 Ω

Solution
Question 10
4 / -1

Curves in the graph shown give, as functions of radial distance r (from the axis), the magnitude B of the magnetic field (due to individual wire) inside and outside four long wires a, b, c and d, carrying currents that are uniformly distributed across the cross sections of the wires. Overlapping portions of the plots are indicated by double labels. All curves start from the origin.

Which wire has the greatest radius ?

A
a

B
b

C
c

D
d

Solution