15 + Interview Questions in JEE MAINS IN GENERAL AWARENESS Page 1 InterviewSolution

1.	Equivalent Resistance between A and B!
Answer» Its a Wheatstone bridge of 4 equal resitamces. So, equivalent resistance will be equal to the either resistance i.e., 10 ohm

Discussion

2.	An ideal liquid of density p is pumped steadily through a tube of radius r at a speed v. The least power of engine required to maintain this flow rate is
Answer» We know that volume rate = Av Volume rate = πr²v dm = πr²v\(\rho\) Power rate = \(\frac12mv^2\) = \(\frac12\)πr²v³\(\rho\)

2.

An ideal liquid of density p is pumped steadily through a tube of radius r at a speed v. The least power of engine required to maintain this flow rate is

Answer»

We know that volume rate = Av

Volume rate = πr²v

dm = πr²v\(\rho\)

Power rate = \(\frac12mv^2\)

= \(\frac12\)πr²v³\(\rho\)

Discussion

3.	PLS EXPLAIN IT PROPERLY
Answer» Answer will be 1:2

Discussion

4.	The minimum energy required to launch a satellite of mass m from the surface of earth of mass M and Radius R in a circular orbit at an altitude of 2R is .
Answer» Correct option is: \(\frac{5\,GMm}{6R}\) Given mass of satellite = M mass of the surface = M Radius = R Altitude h = 2R Gravitational potential energy = \(-\frac{GMm}{r}\) Gravitational potential energy at Altitude = \(-\frac{GMm}{r+h}\) = \(-\frac{GMm}{R+2R}\) = \(-\frac{GMm}{3R}\) Orbital velocity V_o² = \(\frac{GM}{R+h}\) V_o² = \(\frac{GM}{3R}\) total potential energy E = kinetic energy + potential energy E_f = \(\frac{1}{2}\)mv_o² + \((-\frac{GMm}{3R})\) ∵ V_o² = \(\frac{GM}{3R}\) E_f = \(\frac{1}{2}\) \(\frac{GMm}{3R}-\frac{GMm}{3R}\) E_f = \(\frac{GMm-2\,GMm}{6R}\) E_f = \(-\frac{GMm}{6R}\) Ei = Ef the minium energy required = \(\frac{GMm}{R}-\frac{GMm}{6R}\) = \(\frac{6\,GMm-GMm}{6R}\) Minimum energy required = \(\frac{5\,GMm}{6R}\)

4.

The minimum energy required to launch a satellite of mass m from the surface of earth of mass M and Radius R in a circular orbit at an altitude of 2R is .

Answer»

Correct option is: \(\frac{5\,GMm}{6R}\)

Given mass of satellite = M

mass of the surface = M

Radius = R

Altitude h = 2R

Gravitational potential energy = \(-\frac{GMm}{r}\)

Gravitational potential energy at Altitude = \(-\frac{GMm}{r+h}\)

= \(-\frac{GMm}{R+2R}\)

= \(-\frac{GMm}{3R}\)

Orbital velocity V_o² = \(\frac{GM}{R+h}\)

V_o² = \(\frac{GM}{3R}\)

total potential energy E = kinetic energy + potential energy

E_f = \(\frac{1}{2}\)mv_o² + \((-\frac{GMm}{3R})\)

∵ V_o² = \(\frac{GM}{3R}\)

E_f = \(\frac{1}{2}\) \(\frac{GMm}{3R}-\frac{GMm}{3R}\)

E_f = \(\frac{GMm-2\,GMm}{6R}\)

E_f = \(-\frac{GMm}{6R}\)

Ei = Ef

the minium energy required

= \(\frac{GMm}{R}-\frac{GMm}{6R}\)

= \(\frac{6\,GMm-GMm}{6R}\)

Minimum energy required = \(\frac{5\,GMm}{6R}\)

Discussion

5.	WHICH OF THE FOLLOWING ELEMENT DOESN'T SHOW ALLOTROPY? 1) NITROGEN 2) BISMUTH 3) PHOSPHORUS 4) ARSENIC (SINGLE OPTION ONLY)
Answer» Correct answer is:- (1) Nitrogen Nitrogen does not show allotropy due to its weak N-N single bond. Therefore, ability of nitrogen to form polymeric structure is less. Hence, option A is correct.

5.

WHICH OF THE FOLLOWING ELEMENT DOESN'T SHOW ALLOTROPY? 1) NITROGEN 2) BISMUTH 3) PHOSPHORUS 4) ARSENIC (SINGLE OPTION ONLY)

Answer»

Correct answer is:- (1) Nitrogen

Nitrogen does not show allotropy due to its weak N-N single bond. Therefore, ability of nitrogen to form polymeric structure is less. Hence, option A is correct.

Discussion

6.	In complex MCl3.5H2O, the secondary valency of metal is 6 and it has no molecule of water present out of coordination sphere. Calculate the volume of 0.1M AgNO3 solution needed to precipitate the free chloride ions in 200 mL of 0.01 M solution of the complex.(a) 80mL(b) 40mL(c) 20mL(d) 120mLPS- The answer is (b) 40mL
Answer» Correct option is (b) 40ml As we have given, The secondary valency of metal = 6 and There is no molecule of water present out of coordination sphere. Therefore-- Molecular formula of complex will be--- [M(H₂O)₅Cl]Cl₂ In solution 1 mole of [M(H₂O)₅]Cl₂ gives 2 mole of chloride ions. \(\therefore\) 2 mole [M(H₂O)₅Cl]Cl₂ gives = 4 mole chloride ions. Let say V volume of 0.1 M AgNO₃ needed to precipitate free chloride ions present in 200 ml of 0.01 M [M(H₂O)₅Cl]Cl₂. \(\therefore\) 0.1 x V = 2 x(200ml x 0.01) V = 4/0.1 V = 40 ml

6.

In complex MCl3.5H2O, the secondary valency of metal is 6 and it has no molecule of water present out of coordination sphere. Calculate the volume of 0.1M AgNO3 solution needed to precipitate the free chloride ions in 200 mL of 0.01 M solution of the complex.(a) 80mL(b) 40mL(c) 20mL(d) 120mLPS- The answer is (b) 40mL

Answer»

Correct option is (b) 40ml

As we have given,

The secondary valency of metal = 6

and There is no molecule of water present out of coordination sphere.

Therefore--

Molecular formula of complex will be---

[M(H₂O)₅Cl]Cl₂

In solution 1 mole of [M(H₂O)₅]Cl₂ gives 2 mole

of chloride ions.

\(\therefore\) 2 mole [M(H₂O)₅Cl]Cl₂ gives = 4 mole chloride ions.

Let say V volume of 0.1 M AgNO₃ needed to precipitate free chloride ions present in 200 ml of 0.01 M

[M(H₂O)₅Cl]Cl₂.

\(\therefore\) 0.1 x V = 2 x(200ml x 0.01)

V = 4/0.1

V = 40 ml

Discussion

7.	Which of the following demonstrated that earth has a magnetic field:-(A) intensity of cosmic rays (Stream of charged particle coming from outer space) is more at the poles than at the equator(B) Earth is surrounded by an ionsphere (a shell of charged particles)(C) Earth is a planet rotating about the north-south axis(D) large quantity of iron ore is found in the earth
Answer» (A) intensity of cosmic rays (Stream of charged particle coming from outer space) is more at the poles than at the equator.

Discussion

8.	Let α, β and γ be the roots of f(x) = x3 + x2 - 5x - 1 = 0. Then [α] + [β] + [γ], where [.] denotes the greatest integer function, is equal to..........................
Answer» Ans. -1 Sum of 'r'roots of equation=coefficient of x^n-r/coefficient of x^n Where n=highest power of variable Here a=-1 B=1 Y=-1

Discussion

9.	Figure shows (only cross section) a wooden cylinder C with a mass m of 0.25Kg. a radius R and a length l perpendicular to the plane of paper of 0.1 meter with N equal to 10 turns of wire wrapped around it longitudinally, so that the plane of the wire loop contains the axis of cylinder. What is the least current through the loop that will prevent the cylinder from rolling down a plane whose surface is inclined at angle θ to the horizontal, in the presence of a vertical field of magnetic induction 0.5 weber/meter2, if the plane of the windings is parallel to the inclined plane?
Answer» Answer *3 squrt(31) 10^-5 T, 30⁰ E of N**

9.

Figure shows (only cross section) a wooden cylinder C with a mass m of 0.25Kg. a radius R and a length l perpendicular to the plane of paper of 0.1 meter with N equal to 10 turns of wire wrapped around it longitudinally, so that the plane of the wire loop contains the axis of cylinder. What is the least current through the loop that will prevent the cylinder from rolling down a plane whose surface is inclined at angle θ to the horizontal, in the presence of a vertical field of magnetic induction 0.5 weber/meter2, if the plane of the windings is parallel to the inclined plane?

Answer»

Answer

3 squrt(31) *10^-5 T, 30⁰ E of N

Discussion

10.	A compound with molecular mass 180μ is acylated with CH3COCl, to get a compound with molecular mass 390μ. The number of amino groups present per molecule of the former compound is (a) 6 (b) 2 (c) 5 (d) 4
Answer» The correct option is (c) 5. Explanation: Acylation of an amine group is represented as follows: RNH₂ + CH₃COCl ---> RNHCOCH₃ + HCl Mass of product = 390 g Mass of amine = 180 g Increase in mass = 390- 180 g = 210 g Mass of one molecule of CH₃CO = 12 + 3 + 12 + 16 = 43 Since, one hydrogen atom of NH₂ group is replaced mass added per molecule = 43 - 1 = 42 g So, number of aceryl groups introduced = 210/42 = 5 Thus, number of amino groups in the original compound = 5

10.

A compound with molecular mass 180μ is acylated with CH3COCl, to get a compound with molecular mass 390μ. The number of amino groups present per molecule of the former compound is (a) 6 (b) 2 (c) 5 (d) 4

Answer»

The correct option is (c) 5.

Explanation:

Acylation of an amine group is represented as follows:

RNH₂ + CH₃COCl ---> RNHCOCH₃ + HCl

Mass of product = 390 g

Mass of amine = 180 g

Increase in mass = 390- 180 g = 210 g

Mass of one molecule of CH₃CO = 12 + 3 + 12 + 16 = 43

Since, one hydrogen atom of NH₂ group is replaced mass added per molecule = 43 - 1 = 42 g

So, number of aceryl groups introduced = 210/42 = 5

Thus, number of amino groups in the original compound = 5

Discussion

11.	A torque of 1 N.m is applied to a wheel which is at rest. After 2 second the angular momentum in kg.m2/s is :0.52314
Answer» Torque is the rate of change of angular momentum Torque = x−0/t Where x is final angular momentum and initial as the body is at rest, initial angular momentum = 0 x = 1 * 2 = 2

Discussion

12.	If a,b,c are vector such that [a b c ]=4 then value of [a×b b×c c×a] is
Answer» Given [a b c ] = 4 so [ axb bxc cxa]=[ a b c ]²= 4² =16

12.

If a,b,c are vector such that [a b c ]=4 then value of [a×b b×c c×a] is

Answer»

Given [a b c ] = 4

so [ axb bxc cxa]=[ a b c ]²= 4² =16

Discussion

13.	f (x)=3sin4x-cos6x then find the difference of maximum and minimum
Answer» sin(x) and cos(x) both have range [-1,1], it should be obvious that the value of: sin²(x)∈[0,1] sin⁴(x)∈[0,1] 3sin⁴(x)∈[0,3] cos²(x)∈[0,1] cos⁶(x)∈[0,1] The max^m value of 3sin⁴(x) =3 (when x = (2n+1/2)π ) The mini^m value of 3sin⁴(x) = 0 (when x = nπ ) The max^m value of cos⁶(x) =1 (when x = nπ ) The mini^m value of cos⁶(x) = 0 (when x = (2n+1/2)π ) So, at x=nπ, 3sin⁴(x) has its minimum value and cos⁶(x)has its maximum value, so for f(x) to be minimum value (= 0–1 = -1). When x= (2n + 1/2)π, 3sin⁴(x) has its maximum value and cos⁶(x)has its minimum value, so for f(x) to be maximum value (= 3–0 = 3). Hence, the difference between the maximum and minimum values of f(x) = 4

13.

f (x)=3sin4x-cos6x then find the difference of maximum and minimum

Answer»

sin(x) and cos(x) both have range [-1,1], it should be obvious that the value of:

sin²(x)∈[0,1]

sin⁴(x)∈[0,1]

3sin⁴(x)∈[0,3]

cos²(x)∈[0,1]

cos⁶(x)∈[0,1]

The max^m value of 3sin⁴(x) =3 (when x = (2n+1/2)π )
The mini^m value of 3sin⁴(x) = 0 (when x = nπ )

The max^m value of cos⁶(x) =1 (when x = nπ )
The mini^m value of cos⁶(x) = 0 (when x = (2n+1/2)π )

So, at x=nπ, 3sin⁴(x) has its minimum value and cos⁶(x)has its maximum value,
so for f(x) to be minimum value (= 0–1 = -1).

When x= (2n + 1/2)π, 3sin⁴(x) has its maximum value and cos⁶(x)has its minimum value,
so for f(x) to be maximum value (= 3–0 = 3).

Hence, the difference between the maximum and minimum values of f(x) = 4

Discussion

Explore topic-wise InterviewSolutions in .

Equivalent Resistance between A and B!

An ideal liquid of density p is pumped steadily through a tube of radius r at a speed v. The least power of engine required to maintain this flow rate is

PLS EXPLAIN IT PROPERLY

The minimum energy required to launch a satellite of mass m from the surface of earth of mass M and Radius R in a circular orbit at an altitude of 2R is .

WHICH OF THE FOLLOWING ELEMENT DOESN'T SHOW ALLOTROPY? 1) NITROGEN 2) BISMUTH 3) PHOSPHORUS 4) ARSENIC (SINGLE OPTION ONLY)

Let α, β and γ be the roots of f(x) = x3 + x2 - 5x - 1 = 0. Then [α] + [β] + [γ], where [.] denotes the greatest integer function, is equal to..........................

A compound with molecular mass 180μ is acylated with CH3COCl, to get a compound with molecular mass 390μ. The number of amino groups present per molecule of the former compound is (a) 6 (b) 2 (c) 5 (d) 4

A torque of 1 N.m is applied to a wheel which is at rest. After 2 second the angular momentum in kg.m2/s is :0.52314

If a,b,c are vector such that [a b c ]=4 then value of [a×b b×c c×a] is

f (x)=3sin4x-cos6x then find the difference of maximum and minimum