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Question : When one gene pair hides the effect of the other unit, the phenomenon is referred to as:
Option 1: Epistasis
Option 2: Mutation
Option 3: None of the options
Option 4: Dominance
Correct Answer: Epistasis
Solution : The correct option is - Epistasis.
The term "epistasis" refers to a situation in which the expression of one gene is altered (e.g., hidden, inhibited, or suppressed) by the expression of one or more other genes. Epistasis is classified as either recessive or dominant.
Question : The centroid of an equilateral triangle PQR is L. If PQ = 6 cm, the length of PL is :
Option 1: $4 \sqrt{3} \mathrm{~cm}$
Option 2: $3\sqrt{3} \mathrm{~cm}$
Option 3: $2\sqrt{3} \mathrm{~cm}$
Option 4: $5\sqrt{3} \mathrm{~cm}$
Correct Answer: $2\sqrt{3} \mathrm{~cm}$
Solution : PQ = 6 cm The centroid divides the median in the ratio 2 : 1. Since all the centres lie at the same point L, PL is also the circumradius of Equilateral Triangle PQR. We know, Circumradius of an equilateral triangle = $\frac{\text{Side of
Question : Directions: In the following question, a series is given with one (or more) number(s)/alphabet missing. Choose the correct alternative from the given options. J2Z, K4X, L7V, M11T, ?
Option 1: O17R
Option 2: N17S
Option 3: R16N
Option 4: N16R
Correct Answer: N16R
Solution : Given: J2Z, K4X, L7V, M11T, ?
Add 1 to the place value of the first letter and add consecutive natural numbers starting from 2 to the number of the previous term respectively, and subtract 2 from the third letter to obtain the next term in
Question : Which of the following battles led to the foundation of the Mughal rule at Delhi ?
Option 1: Third battle of Panipat
Option 2: Second battle of Panipat
Option 3: Battle of Haldighati
Option 4: First battle of Panipat
Correct Answer: First battle of Panipat
Solution : The correct option is First battle of Panipat.
The Sultan of Delhi, Ibrahim Lodi, and the Mughal army, led by Babur, engaged in the battle that took place on April 20, 1526, near the town of Panipat in present-day Haryana. Sultan Ibrahim
Question : Directions: In the following question, some parts of the sentence have errors, and some are correct. Find out which part of the sentence has an error. The number of that part is the answer. If a sentence is error-free, your answer is "No Error".
Most of the members at the meeting felt (1)/ that the group appointed to investigate the case (2)/ were not competent to do the job efficiently. (3)/ No Error (4)
Option 1: (1)
Option 2: (2)
Option 3: (3)
Option 4: (4)
Correct Answer: (3)
Solution : The error lies in the third part of the sentence.
The sentence contains an error in the subject-verb agreement. The subject group is singular, so the verb should also be singular.
Were is the incorrect form of the verb; it should be was.
Therefore, the
Question : A tangent AB at point A of a circle of radius 6 cm meets a line through the centre O at point B. If OB = 10 cm, then the length of AB (in cm) is equal to:
Option 1: 5
Option 2: 6
Option 3: 4
Option 4: 8
Correct Answer: 8
Solution : Given, AB is a tangent, OB = 10 cm and OA = 6 cm Since tangent is perpendicular to the radius at its point of contact, $\angle$OAB = 90$^\circ$ Using Pythagoras theorem, OB2 = AB2 + OA2 ⇒ 102 = AB
Question : Directions: A series is given below with one term missing. Choose the correct alternative from the given ones that will complete the series. VW, XU, ZS, BQ, ?
Option 1: OD
Option 2: DO
Option 3: EO
Option 4: FO
Correct Answer: DO
Solution : Given: VW, XU, ZS, BQ, ?
Add 2 to the positional value of the first letter and subtract 2 from the positional value of the second letter of the previous terms to get the first and second letters respectively of the next term. First letter
Question : If $\cot \theta=\frac{1}{\sqrt{3}}, 0^{\circ}<\theta<90^{\circ}$, then the value of $\frac{2-\sin ^2 \theta}{1-\cos ^2 \theta}+\left(\operatorname{cosec}^2 \theta-\sec \theta\right)$ is:
Option 1: 0
Option 2: 2
Option 3: 5
Option 4: 1
Correct Answer: 1
Solution : $\cot \theta=\frac{1}{\sqrt{3}}$ ⇒ $\theta=60^{\circ}$ So, $\frac{2-\sin ^2 \theta}{1-\cos ^2 \theta}+\left(\operatorname{cosec}^2 \theta-\sec \theta\right)$ = $\frac{2-\sin ^2 60^{\circ}}{1-\cos ^2 60^{\circ}}+\left(\operatorname{cosec}^2 60^{\circ} -\sec 60^{\circ}\right)$ = $\frac{2- \frac{3}{4}}{1-\frac{1}{4}}+\frac{4}{3} -2$ = $\frac{\frac{5}{4}}{\frac{3}{4}}+\frac{4}{3} -2$ = $\frac{5}{3}+\frac{4}{3} -2$ = $\frac{9}{3} -2$ = 3 – 2 = 1 Hence, the correct answer is
Question : ABCD is a cyclic quadrilateral and BC is the diameter of the related circle on which A and D also lie. $\angle \mathrm{BCA}=19°$ and $\angle \mathrm{CAD}=32°$. What is the measure of $\angle \mathrm{ACD}$?
Option 1: 41°
Option 2: 38°
Option 3: 40°
Option 4: 39°
Correct Answer: 39°
Solution : Given, $\angle{CAD}=32°$ and $\angle{BCA}=19°$ We know, that the angle formed by the diameter is 90°. ⇒ $\angle{BAC}=90°$ The sum of the opposite angle of a quadrilateral = 180°. ⇒ $\angle{CAD}+\angle{BAC}+\angle{BCA}+\angle{ACD}=180°$ ⇒ $32°+90°+19°+\angle{ACD}=180°$ ⇒ $141°+\angle{ACD}=180°$ ⇒ $\angle{ACD}=180°-141°$ ⇒ $\angle{ACD}=39°$ Hence, the correct answer is 39°.
Question : There were several cities in the Maurya empire. Which of the following cities was located on the southernmost side?
Option 1: Girnar
Option 2: Topra
Option 3: Brahmagiri
Option 4: Rupnath
Correct Answer: Brahmagiri
Solution : The correct option is Brahmagiri.
Brahmagiri is an archaeological site located in the Chitradurga district of Karnataka. Benjamin L. Rice discovered rock edicts of Emperor Ashoka on this site in 1891. These rock edicts of the area assure that it was the southernmost part
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