Research Aptitude Test

Result

Research Aptitude Test - 27

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
2 / -0.5

A known positive charge is located at point P between two unknown charges, Q1 and Q2. P is closer to Q2 than Q1. If the net electric force acting on the charge at P is zero, it may correctly be concluded that:

A
Both Q1 and Q2 are positive

B
Both Q1and Q2 are negative

C
Q1 and Q2 have opposite signs

D
Q1 and Q2 have the same sign, but magnitude of Q1 is greater than the magnitude of Q2

Solution

To Balance the Forces, it is clear that Q1, Q2 must have the same sign of charges by which Both will Attract or Both will Repel to Balance the forces.

As Distance is not Same so, Their Magnitude be different. F = kq₁ q₂ /d as F ~ 1/d (Inversely Proportion) so in between Q2 &P as d is Less Hence F is more to Balance it so, Q2 must be more.
Question 2
2 / -0.5

Which of the following solutions, when mixed, will not form a buffer solution?

A
A. 100mL 0.1M NaOH + 50mL 0.1M CH₃ COOH

B
B. 50mL 0.1M NaOH + 100mL 0.1M CH₃ COOH

C
C. 50mL 0.1M NH₄ OH + 50mL 0.1M CH₃ COOH

D
D. 50mL 0.1M HCl + 100mL 0.1M CH₃ COONa

Solution

Buffer Solution is a Solution which Resist change in pH

For a Buffer Solution: Existence of both weak part & its Conjugate is Required.

In (A) As NaOH is more than CH₃ COOH so, CH₃ COOH will be the LR (Limiting Reagent) & No Conjugate Part Will Remain.
Question 3
2 / -0.5

Three identical masses are at the three corners of the triangle, connected by massless identical springs (rest length 10) forming an isosceles right-angle triangle. If the two sides of equal length (of length 210) lie along positive x-axis and positive y-axis, then the force on the mass that is not at the origin but on the x-axis is given by with

A
A. a = 1 and b = 0

B
B. a = 0 and b = 1

C
C. a = -√2 and b = 1

D
D. a = - 2 and b = 0

E
E. a = - 2 and b = 1

Solution

Que Directly from NCERT Exemplar Available on IIITprep.com SUPR Book

Mainly 2 types of forces are Considered here

1. Gravitational Force

2. Spring Force

The Spring force is kx² where x is Elongation. Now take the Proper Direction of Forces. Resolve its Component along x axis & y axis i.e. so, a = -2 & b = 1
Question 4
2 / -0.5

Consider a rope fixed at both ends under tension so that it is horizontal (i.e., assume the rope is along the x-axis, with gravity acting along the z-axis). Now the right end is continually oscillated at high frequency n (say n=100 Hz) horizontally and in a direction along the rope; amplitude of oscillation is negligible. The oscillation travels along the rope and is reflected at the left end.

Let the total length of rope be I, total mass be m and the acceleration due to gravity be g.

After initial phase (say a minute or so), the rope has (BLANK-1)_ wave, which is_(BLANK-2)__in nature. It results from superposition of left travelling and right traveling _(BLANK-3) waves. This resulting wave has a frequency__(BLANK-4) _ that of oscillation frequency nu, Simpy &fmensional analysis indicates that the frequency of can be of the form:___(BLANK-5)__.

A
stationary, transverse, regular, equal to, sqrt (g/l)

B
stationary, , regular, equal to, sqrt (g/l), transverse

C
transverse, regular, equal to, sqrt (g/l), stationary

D
stationary, regular, sqrt (g/l), transverse, equal to

E
stationary, sqrt (g/l), transverse, equal to, regular

Solution

Answer:

1. Nature of the resulting wave
The resulting wave on the rope after the initial phase is a transverse wave.

2. Frequency of the resulting wave
The frequency of the resulting wave is equal to the oscillation frequency n .

3. Superposition of waves
The resulting wave is formed due to the superposition of left traveling and right traveling waves.

4. Form of the frequency of the resulting wave
The frequency of the resulting wave can be expressed in the form:

(sqrt(g/l))

Explanation:

Let 's analyze the given scenario step by step to understand the nature of the wave and its frequency.

1. As the right end of the rope is oscillated horizontally at a high frequency n, a wave is generated that travels along the rope. This wave can be considered as a right traveling wave.

2. When this wave reaches the left end of the rope, it gets reflected. The reflected wave can be considered as a left traveling wave.

3. The resulting wave on the rope is formed due to the superposition of the left and right traveling waves. Since the rope is fixed at both ends, the resulting wave is a standing wave.

4. The standing wave on the rope has a frequency that is equal to the oscillation frequency n. This means that the frequency of the resulting wave is equal to the frequency of the oscillation applied at the right end of the rope.

5. Dimensional analysis suggests that the frequency of the resulting wave can be expressed in terms of the acceleration due to gravity (g) and the length of the rope (l). The only combination that gives a frequency is sqrt(g/l).

Therefore, the answer is: stationary, transverse, regular, equal to, sqrt (g/l)
Question 5
2 / -0.5

The value of (√8)^⅓is:

A
A. 2

B
B. 4

C
C. √2

D
D. 8

Solution
Question 6
2 / -0.5

(1/216)^-⅔ / (1/27)^-4/3 = ?

A
3/4

B
2/3

C
4/9

D
1/8

Solution
Question 7
2 / -0.5

If √2ⁿ= 64, then the value of n is:

A
A. 8

B
B. 4

C
C. 6

D
D. 12

Solution

√2ⁿ = 64 ⇒ 2^n/2 = 64 = 2⁶ .

∴ n/2 = 6 or n = 12
Question 8
2 / -0.5

If then n equals:

A
0

B
2

C
3

D
4

Solution

Or 3^{2n + 14} = 3²⁰

∴2n + 14 = 20

Or 2n = 6 or n = 3.
Question 9
2 / -0.5

How many terms are there in the AP 20, 25, 30, .............. 130.

A
A. 22

B
B. 23

C
C. 21

D
D. 24

Solution

In series 20, 25, 3 0 .............. 130.

a = 20, d = 5

n^th term is 130

⇒ 20 + (n - 1) 5 = 130

⇒ 5n = 115, n = 23
Question 10
2 / -0.5

Find the 1st term of an AP whose 8th and 12 m terms are respectively 39 and 59.

A
A. 5

B
B. 6

C
C. 4

D
D. 3

Solution

Here a + 7d = 39 …(i)

a + 11d = 59

So, 4d = 20

d = 5

a + (7 x 5) = 39

So, a = 4