987 + Interview Questions in KVPY IN GENERAL AWARENESS Page 10 InterviewSolution

451.	A new life from discovered on a distant planet has a genetic code consisting of five unique nucleotides and only one stop codon. If each has four bases, what is the maximum number of unique amino acids this life from can use?A. 624B. 20C. 124D. 3124
Answer» Correct Answer - A No. of unique nucleotide =5 N0. of bases in codon = 4 So, total combiations `rArr5^(4)rArr625` No. of stop condons =1 so, unique amino acids (maximum ) `=625-1rArr624`

Discussion

452.	Let `nge3`. A list of numbers `x_(1),x_(2),....,x_(n)` has mean `mu` and standard deviation `sigma` . A new list of numbers `y_(1),(y_(2),....,y_(n)` is made as follows `: y_(1)=(x_(1)+ x_(2))/(2),y_(2)=(x_(1)+x_(2))/(2) and y_(j)"for"j=3,4,....,n`. The mean and the standard deviation of the new list are `hatmuand hatsigma` . Then whcih of the following is necessarily true?A. `mu=hatmu and sigmalehat sigma`B. `mu=hatmu and sigma ge hat sigma`C. `sigma-hat sigma`D. `munehat mu`
Answer» Correct Answer - B `mu=(x _(1)+x_(2)+.....x_(n))/(n)` `hatmu=(y_(1)+y_(2)+....+y_(n))/(n)=((x_(1)+x_(2))/(2)+(x_(1)+x_(2))/(2)+x_(3)+....+x_(n))/(n)` `hatmu=(x_(1)+x_(2)+....+x_(n))/(n)=murArrhatmu=mu` `sigma^(2)=(sumx_(i)^(2))/(n)-mu^(2)` `sigma^(2)=(x_(1)^(2)+x_(2)^(2)+....+x_(n)^(2))/(n)-mu^(2)" "....(1)` `hatsigma^(2)=(x_(1)^(2)+y_(2)^(2)+....+y_(n)^(2))/(n)-mu^(2) (((x_(1)+x_(2))/(2))^(2)+((x_(1)+x_(2))/(2))^(2)+x_(3)^(2)+....x+_(n)^(2))/(n)-mu^(2)` `hatsigma^(2)=((x_(1)^(2)+x_(2)^(2))/(2)+x_(1)x_(2)+x_(3)^(2)+....+x_(n)^(2))/(n)-mu^(2)" " ....(2)` `sigma^(2)-hatsigma^(2)=(x_(1)^(2)+x_(2)^(2))/(n)-((x_(1)^(2)+x_(2)^(2)+2x_(1)x_(2))/(2n))=(x_(1)^(2)+x_(2)^(3)-2x_(1)x_(2))/(2n)` `(x _(1)-x_(2))^(2)/(2n)ge0rArrgehat sigma&mu=hatmu`

Discussion

453.	Following the cell cycle scheme given below, what is the probaility that a cell would be in M-phase at any given time ? A. `1//2`B. `1//12`C. `1//6`D. `1//2`
Answer» Correct Answer - B Total time for cell cycle =24 hrs. Time for M-phase =2 hrs So, probability of cell in M-phase at any given time is `(2)/(24)rarr(1)/(12)`

Discussion

454.	Consider the quadratic equation `nx^(2) +7 sqrt(nx)+n=0`, where n is a positive intergar. Which of the following statements are necessarliy correct ? I. For any n, the roots are distinct. (II) There are infinitely many values of n for which both roots are real. (III) The product of the roots is necessarlity an integer.A. III onlyB. I and II onlyC. II and III onlyD. I, II and III
Answer» Correct Answer - b `D=49n-4n^(2)` `=n(49-4n)` `Dne0 "for any n" in I^(+)` so roots are distinct. for roots to be real `D ge 0` So `n le49/4` so n can be { 1,2,3 â€¦â€¦â€¦.12 } Clearly product of the roots is 1.

Discussion

455.	If the axis of rotation of the earth were extended into space then it would pass close to -A. the moonB. the sunC. the pole starD. the centre of mass of all the planets in the solar system
Answer» Correct Answer - C

Discussion

456.	Figure below shows a portion of an electric circuit with the currents in amperes and their directions. The magnitude and direction of the current in the portion PQ is A. 0AB. 3 A from P to QC. 4 A from Q to PD. 6A from Q to P
Answer» Correct Answer - D

Discussion

457.	A hollow lens is made of thin glass and in the shape of a double concave lens. It can be filled with air, water of refractive index `1.33` or `CS_(2)` of refractive index `1.6`. It will act as a diverging lens if it isA. filled with air and immersed in water.B. filled with water and immersed in`CS_(2)`.C. filled with air and immersed in `CS_(2)`.D. filled with `CS_(2)` and immersed in water.
Answer» Correct Answer - D

Discussion

458.	A loop carrying current I has the shape of a regular polygon of n sides. If R is the distance from the centre to any vertex, then the magnitude of the magnetic induction vector `vec(B)` at the centre of the loop is –A. `n(mu_(0)I)/(2pi R)tan.(pi)/(n)`B. `(mu_(0)I)/(2R)`C. `n(mu_(0)I)/(2pi R)tan.(2pi)/(n)`D. `(mu_(0)I)/(pi R)tan.(pi)/(n)`
Answer» Correct Answer - A

Discussion

459.	In a thermally isolated system, two boxes filled with an ideal gas are connected by a valve. When the valve is in closed position, states of the box 1 and 2, respectively, are (1 atm, V, T) and (0.5 atm, 4V, T). When the valve is opened, the final pressure of the system is approximatelyA. 0.5 atmB. 0.6 atmC. 0.75 atmD. 1.0 atm
Answer» Correct Answer - B

Discussion

460.	A body is executing simple harmonic motion of amplitude a and period T about the equilibrium position x = 0. large numbers of snapshots are taken at random of this body in motion. The probability of the body being found in a very small interval x to x + \|dx\| is highest at – (A) x = ± a (B) x = ± a/2 (C) x = 0 (D) x = ± / √2
Answer» Correct :- (A) x = ± a

460.

A body is executing simple harmonic motion of amplitude a and period T about the equilibrium position x = 0. large numbers of snapshots are taken at random of this body in motion. The probability of the body being found in a very small interval x to x + |dx| is highest at – (A) x = ± a (B) x = ± a/2 (C) x = 0 (D) x = ± / √2

Answer»

Correct :- (A) x = ± a

Discussion

461.	A conducting rod of mass m and length l is free to move without friction on two parallel long conducting rails, as shown below. There is a resistance R across the rails. In the entire space around, there is a uniform magnetic field B normal to the plane of the rod and rails. The rod is given an impulsive velocity v0 – Finally, the initial energy 1/2 mv20(A) Will be converted fully into heat energy in the resistor (B) Will enable rod to continue to move with velocity v0 since the rails are frictionless. (C) Will be converted fully into magnetic energy due to induced current (D) Will be converted into the work done against the magnetic field
Answer» Correct :- (A) Will be converted fully into heat energy in the resistor Explanation :- Due to energy conservation

461.

A conducting rod of mass m and length l is free to move without friction on two parallel long conducting rails, as shown below. There is a resistance R across the rails. In the entire space around, there is a uniform magnetic field B normal to the plane of the rod and rails. The rod is given an impulsive velocity v0 – Finally, the initial energy 1/2 mv20(A) Will be converted fully into heat energy in the resistor (B) Will enable rod to continue to move with velocity v0 since the rails are frictionless. (C) Will be converted fully into magnetic energy due to induced current (D) Will be converted into the work done against the magnetic field

Answer»

Correct :- (A) Will be converted fully into heat energy in the resistor

Explanation :-

Due to energy conservation

Discussion

462.	A body is executing simple harmonic motion of amplitude a and period T about the equilibrium position x = 0. large numbers of snapshots are taken at random of this body in motion. The probability of the body being found in a very small interval x to `x + \|dx\|` is highest at –A. `x = +- a`B. `x = +- a//2`C. `x = 0`D. `x = +-//sqrt(2)`
Answer» Correct Answer - A

Discussion

463.	A steady current I flows through a wire of radius r, length L and resistivity ρ. The current produced heat in the wire. The rate of heat loss in a wire is proportional to its surface area. The steady temperature of the wire is independent of– (A) L (B) r (C) I (D) ρ
Answer» Correct :- (A) L Explanation :- Concept of fuse wire

463.

A steady current I flows through a wire of radius r, length L and resistivity ρ. The current produced heat in the wire. The rate of heat loss in a wire is proportional to its surface area. The steady temperature of the wire is independent of– (A) L (B) r (C) I (D) ρ

Answer»

Correct :- (A) L

Explanation :-

Concept of fuse wire

Discussion

464.	A conducting rod of mass m and length l is free to move without friction on two parallel long conducting rails, as shown below. There is a resistance R across the rails. In the entire space around, there is a uniform magnetic field B normal to the plane of the rod and rails. The rod is given an impulsive velocity `v_(0)`- Finally, the initial energy `(1)/(2) mv_(0)^(2)`A. Will be converted fully into heat energy in the resistorB. Will enable rod to continue to move with velocity `v_(0)` since the rails are frictionlessC. Will be converted fully into magnetic energy due to induced currentD. Will be converted into the work done against the magnetic field
Answer» Correct Answer - A

Discussion

465.	The ratio of the speed of sound to the average speed of an air molecule at 300 K and 1 atmospheric pressure is close to–A. 1B. `sqrt(1//300)`C. `sqrt(300)`D. 300
Answer» Correct Answer - A

Discussion

466.	A steady current I flows through a wire of radius r, length L and resistivity `rho`. The current produced heat in the wire. The rate of heat loss in a wire is proportional to its surface area. The steady temperature of the wire is independent of–A. LB. rC. ID. `rho`
Answer» Correct Answer - A

Discussion

467.	The maximum number of electrons that can be filled in the shell with the principal quantum number n = 4 is (A) 64 (B) 26 (C) 18 (D) 32
Answer» Correct Option :- (D) 32 Explanation :- 4s, 4p, 4d & 4f contains total 32 electrons.

467.

The maximum number of electrons that can be filled in the shell with the principal quantum number n = 4 is (A) 64 (B) 26 (C) 18 (D) 32

Answer»

Correct Option :- (D) 32

Explanation :-

4s, 4p, 4d & 4f contains total 32 electrons.

Discussion

468.	White phosphorous catches fire in air to produce dense white fumes. This is due to the formation of (A) P4O10 (B) PH3 (C) H3PO3 (D)H3 PO2
Answer» Correct Option :- (A) P₄O₁₀

468.

White phosphorous catches fire in air to produce dense white fumes. This is due to the formation of (A) P4O10 (B) PH3 (C) H3PO3 (D)H3 PO2

Answer»

Correct Option :- (A) P₄O₁₀

Discussion

469.	White phosphorous catches fire in air to produce dense white fumes. This is due to the formation ofA. `P_(4)O_(10)`B. `PH_(3)`C. `H_(3)PO_(3)`D. `H_(3)PO_(2)`
Answer» Correct Answer - A Theoritical

Discussion

470.	Two particles of identical mass are moving in circular orbits under a potential given by `V(r) = Kr^(-n)`, where K is a constant. If the radii of their orbits are `r_(1), r_(2)` and their speeds are `v_(1), v_(2)`, respectively, thenA. `v_(1)^(2)r_(1)^(n)=v_(2)^(2)r_(2)^(n)`B. `v_(1)^(2)r_(1)^(-n)=v_(2)^(2)r_(2)^(-2)`C. `v_(1)^(2)r_(1)=v_(2)^(2)r_(2)`D. `v_(1)^(2)r_(1)^(2-n)=v_(2)^(2)r_(2)^(2-n)`
Answer» Correct Answer - A

Discussion

471.	A person goes from point P to point Q covergin 1/3 of the distance with speed 10km/hr, the next 1/3 of the distance at 20 km/hr and the last 1/3 of the distance at 60 km/hr. The average speed of the person isA. 30 km/hrB. 24 km/hrC. 18 km/hrD. 12 km/hr
Answer» Correct Answer - C

Discussion

472.	Thendrills are modifications of ?A. Stem or leafB. Stem onlyC. Leaf onlyD. Aerial roots only
Answer» Correct Answer - A

Discussion

473.	Erythropoietin is produced by (A) heart (B) kidney (C) bone marrow (D) adrenal gland
Answer» Correct option (B) kidney Explanation: Justa glomerular cells of kidney produce erythroprotein hormone during oxygen deficiency. It promotes red bone marrow to form RBC (Erythropoisis).

473.

Erythropoietin is produced by (A) heart (B) kidney (C) bone marrow (D) adrenal gland

Answer»

Correct option (B) kidney

Explanation:

Justa glomerular cells of kidney produce erythroprotein hormone during oxygen deficiency. It promotes red bone marrow to form RBC (Erythropoisis).

Discussion

474.	Action potential in neurons is generated by a rapid influx of (A) chloride ions (B) potassium ions (C) calcium ions (D) sodium ions
Answer» Correct option (D) sodium ions Explanation: Sodium an influx into the axoplasm (three sodium ion) that stimulate neural conduction.

474.

Action potential in neurons is generated by a rapid influx of (A) chloride ions (B) potassium ions (C) calcium ions (D) sodium ions

Answer»

Correct option (D) sodium ions

Explanation:

Sodium an influx into the axoplasm (three sodium ion) that stimulate neural conduction.

Discussion

475.	The water potential `(Psi^(P))` of pure water at standard temperature and atmospheric pressure is ?
Answer» Correct Answer - A

Discussion

476.	A couple has two sons and two daughters. Only one son is colour blind and the rest of the siblings are normal. Assuming colour blindness is sex-linked, which ONE of the following would be the phenotype of the parents ?A. Mother would be colour blind, father would be normal.B. Father would be colour blind, mother would be normal.C. Both the parents would be normal.D. Both the parents would be colour blind.
Answer» Correct Answer - C Male child recieve X-chromosome from mother only. Another normal son indicates that mother is carrier.

Discussion

477.	The number of rectangles that can be obtained by joining four of the twelve vertices of a 12-sided regular polygon is- (A) 66 (B) 30 (C) 24 (D) 15
Answer» Correct Option :-(D) 15 Explanation :- Number of diagonals passing through centre = 6 number of rectangles = ⁶C₂ = 15

477.

The number of rectangles that can be obtained by joining four of the twelve vertices of a 12-sided regular polygon is- (A) 66 (B) 30 (C) 24 (D) 15

Answer»

Correct Option :-(D) 15

Explanation :-

Number of diagonals passing through centre = 6

number of rectangles = ⁶C₂ = 15

Discussion

478.	The locus of the point P = (a, b) where a, b are real numbers such that the roots of `x^(3)+ax^(2)+bx+a =0` are in arithmetic progression is -A. an ellipseB. a circleC. a parabola whose vertex in on the y-axisD. a parabola whose vertex is no the x-axis
Answer» Correct Answer - C `{:("Let roots "alpha-d", " alpha", " alpha+d,,"product"),("Sum "3alpha=-a rArr alpha=-a/3,,alpha(alpha^(2)-d^(2))=-a),("pair product "b=alpha^(2)-alphad+alpha^(2)+alphad+alpha^(2)-d^(2),,alpha^(2)-d^(2)=3):}` `b=2a^(2)+3` `b-3=2/9 a^(2) rArr" locus "x^(2)=9/2 (y-3)` parabola

Discussion

479.	Suppose the sum of the dirst m teams of a arithmetic progression is n and the sum of its first n terms is m, where `mnen.` Then the dum of the first (m + n) terms of the arithmetic progression isA. 1- mnB. mn- 5C. `-(m = n)`D. m + n
Answer» Correct Answer - C `S_(m)=m/2[2a+(m-1)d]=n" …(1)"` `S_(n)=n/2[2a+(n-1)d]=m" …(2)"` by (1) and (2) `(m-n)a+(m-n){m+n-1}d/2= -(m-n)` `rArr2a+(m+n-1)d= -2(mnen)` `rArrS_(m+n)=(m+n)/2[2a+(m+n-1)d]= -(m+n)`

Discussion

480.	If n is a positive integer and `omegane1` is a cube of unity, the number of possible values of `\|e^(sum_(k=0)^(n) ((n)/(k))omega^(k))\|`A. 2B. 3C. 4D. 6
Answer» Correct Answer - C `underset(k-0)overset(n)sum .^(n)C_(k)omega^(k)=""^(n)C_(1)omega+....+""^(c)C_(n)omega^(n)` `=(1omega)^(n)=(-omega^(2))^(n)` `(-1)^(n) omega^(2n)` `:. \|e^((-1)^(n)omega^(2n))\|=\|e^((-omega^(2))^(n))\|` `=\|e^(-cos.(npi)/(3)isin""(4pi)/(3))\|` `=\|e^(cos.(npi)/(3))\|` can have values `={e^(1),e^(1//2),e^(-1//2),e^(-1)}` Four values.

Discussion

481.	Let f : [0, 1] → [–1, 1] and g : [–1, 1] → [0, 2] be two functions such that g is injective and gof : [0, 1] → [0, 2] is surjective. Then (A) f must be injective but need not be surjective (B) f must be surjective but need not be injective (C) f must be bijective (D) f must be a constant function
Answer» Correct option (B) f must be surjective but need not be injective Explanation: f : [0, 1] → [–1, 1] g : [1, 1] → [–0, 2] g is injective & gof is surjective. ⇒ f must be surjective otherwise f(x) would not cover the whole co-domain [–1, 1] (which is also the domain of g) & then consecutively gof would not be able to cover the whole [0, 2] (as g is injective).

481.

Let f : [0, 1] → [–1, 1] and g : [–1, 1] → [0, 2] be two functions such that g is injective and gof : [0, 1] → [0, 2] is surjective. Then (A) f must be injective but need not be surjective (B) f must be surjective but need not be injective (C) f must be bijective (D) f must be a constant function

Answer»

Correct option (B) f must be surjective but need not be injective

Explanation:

f : [0, 1] → [–1, 1]

g : [1, 1] → [–0, 2]

g is injective & gof is surjective.

⇒ f must be surjective otherwise f(x) would not cover the whole co-domain [–1, 1] (which is also the domain of g) & then consecutively gof would not be able to cover the whole [0, 2] (as g is injective).

Discussion

482.	Match the vitamins listed in Column - I with their respective coenzyme form in Column - II . Choose the correct combination . A. P-v , Q - iii , R-`i` , S-ivB. P-iii , Q - iv , R - ii , S- `i`C. P-`i` , Q - ii , R - v, S - iiiD. P-`i` , Q - iv , R - ii , S - iii
Answer» Correct Answer - C

Discussion

483.	A lysosome vesicle of 1 `mu` m diameter has an internal pH of `5.0` . The total number of `H^(+)`ions inside this vesicle would range fromA. `10^(3)` to `10^(4)`B. `10^(4)` to `10^(5)`C. `10^(5)` to `10^(10)`D. `10^(10)` to `6.023 xx 10^(23)`
Answer» Correct Answer - A

Discussion

484.	If the acidic, basic and hydrophobic residues of proteins are considered to be red, green and blue in color, respectively, then a globular protein in aqueous solution would have. (A) red and blue on the surface and green at the core (B) red and green on the surface and blue at the core (C) blue on the surface and red ad green at the core (D) blue and green on the surface and red at the core
Answer» The Correct option is (B) red and green on the surface and blue at the core

Discussion

485.	Let S be the set of all ordered pairs (x,y) of positive intergers, with HCF (x,y)=16 and LCM (x,y) = 48000. The number of elements in S isA. 4B. 8C. 16D. 32
Answer» Correct Answer - B `48000=16xx3000` `=16xx[3^(1)x2^(3)xx5^(3)]` As H.C.F is 16 So `2^(3)` can be selected in 1 way `& 3^(1)xx5^(3) "can be seleted in "(1+1)(3+1)=8 " ways "` No of ordred pairs = 8

Discussion

486.	Consider a three-digit number with the following properties: I. If its digits in units place and tens place are interchanged, the number increases by 36; II. If its digits in units place and hundreds place are interchanged, the number decreases by 198. Now suppose that the digits in tens place and hundreds place are interchanged. Then the number(A) increases by 180 (B) decreases by 270 (C) increases by 360 (D) decreases by 540
Answer» Correct Option :- (D) decreases by 540 Explanation :- Let Three digit No is 100 a + 10 b + c Given 100 a + 10 b + c = 100 a + 10 c + b – 36 9b – 9c + 36 = 0 c = b + 4 b = c – 4 ...(i) Also given 100 a + 10 b + c = 100c + 10 b + a + 198 99 a – 99 c = 198 a = c + 2 ...(ii) ∴ Now 100 a + 10 b + c – (100 b + 10 a + c) = 90 (a – b) = 90 ( c + 2 – c + 4) (use (i) & (ii) ) = 540 ∴ value decrease by 540

486.

Consider a three-digit number with the following properties: I. If its digits in units place and tens place are interchanged, the number increases by 36; II. If its digits in units place and hundreds place are interchanged, the number decreases by 198. Now suppose that the digits in tens place and hundreds place are interchanged. Then the number(A) increases by 180 (B) decreases by 270 (C) increases by 360 (D) decreases by 540

Answer»

Correct Option :- (D) decreases by 540

Explanation :-

Let Three digit No is 100 a + 10 b + c

Given 100 a + 10 b + c = 100 a + 10 c + b – 36

9b – 9c + 36 = 0

c = b + 4

b = c – 4 ...(i)

Also given 100 a + 10 b + c = 100c + 10 b + a + 198

99 a – 99 c = 198

a = c + 2 ...(ii)

∴ Now 100 a + 10 b + c – (100 b + 10 a + c)

= 90 (a – b)

= 90 ( c + 2 – c + 4) (use (i) & (ii) )

= 540

∴ value decrease by 540

Discussion

487.	Consider four triangles having sides (5,12,9), (5,12,11), (5,12,13) and (5,12,15). Among these, the triangle having maximum area has sidesA. (5,12,9)B. (5,12,11)C. (5,12,13)D. (5,12,15)
Answer» Correct Answer - C `"Clearly area of "triangle "having sides "(5,12,13)` `"is greatest (use"triangle=sqrt(s(s-a)(s-b)(s-c))`

Discussion

488.	In a classroom, one-fifth of the boys leave the class and the ratio of the remaining boys to girls is 2:3. If further 44 girls leave the class, the ratio of boys to girls is 5:2. How many more boys should leave the class so that the number of boys equals that of girls ?A. 16B. 24C. 30D. 36
Answer» Correct Answer - B Let no of Boys = x `&` Let no of girls = y `"Given "(((4x)/(5)))/(y)=(2)/(3)` `(2x)/(5y)=(1)/(3)` `y=(6x)/(5)` `"Also, "(((4x)/(5)))/(y-44)=(5)/(2)` `8x=25((6x)/(5)-44)" ""(use 1)"` x=50 y=60

Discussion

489.	Consider four triangles having sides (5,12,9), (5,12,11), (5,12,13) and (5,12,15). Among these, the triangle having maximum area has sides(A) (5,12,9) (B) (5,12,11) (C) (5,12,13) (D) (5,12,15)
Answer» Correct Option :- (C) (5,12,13) Explanation :- Clearly area of Δ having sides (5, 12, 13) is greatest (use Δ = √s(s – a)(s – b)(s – c))

489.

Consider four triangles having sides (5,12,9), (5,12,11), (5,12,13) and (5,12,15). Among these, the triangle having maximum area has sides(A) (5,12,9) (B) (5,12,11) (C) (5,12,13) (D) (5,12,15)

Answer»

Correct Option :- (C) (5,12,13)

Explanation :-

Clearly area of Δ having sides (5, 12, 13) is greatest (use Δ = √s(s – a)(s – b)(s – c))

Discussion

490.	Consider a three-digit number with the following properties: I. If its digits in units place and tens place are interchanged, the number increases by 36, II. If its digits in units place and hundreds place are interchanged, the number decreases by 198. Now suppose that the digits in tens place and hundreds place are interchanged. Then the number.A. increases by 180B. decreases by 270C. increases by 360D. decreases by 540
Answer» Correct Answer - D Let Three digit No is 100 a + 10 b + c Given 100 a + 10 b + c =100 a + 10 c + b-36 9b-9c+36=0 c=b+4 b=c-4 Also given 100 a + 10 b +c = 100c+ 10 b + a + 198 99 a - 99 c = 198 a = c + 2 `therefore` Now 100 a + 10 b + c - (100 b+ 10 a + c) `=90 (a-b)` `= 90 (c+2-c+4) " "(use (i) & (ii))` =540 `therefore` value decreases by 540

Discussion

491.	What is the probability that a human individual would receive the entire haploid set of chromsomes from his her grandfather ?A. `1//2`B. `(1//2)^(23)`C. `(1//2)^(2)`D. `(1//2)^(46)`
Answer» Correct Answer - B

Discussion

492.	In which phase of the cell cycle are sister chromatids available as template for repair ? (A) G1 phase (B) G2 phase (C) S phase (D) M phase
Answer» Correct Option :- (D) M phase Explanation :- Because in M phase sister chromatid form.

492.

In which phase of the cell cycle are sister chromatids available as template for repair ? (A) G1 phase (B) G2 phase (C) S phase (D) M phase

Answer»

Correct Option :- (D) M phase

Explanation :-

Because in M phase sister chromatid form.

Discussion

493.	The major constituents of neurofilaments are -A. microtubulesB. intermediate filamentsC. actin filamentsD. protofilaments
Answer» Correct Answer - D

Discussion

494.	The major constituents of neurofilaments are- (A) microtubules (B) intermediate filaments (C) actin filaments (D) protofilaments
Answer» Correct Option :- (D) protofilaments

494.

The major constituents of neurofilaments are- (A) microtubules (B) intermediate filaments (C) actin filaments (D) protofilaments

Answer»

Correct Option :- (D) protofilaments

Discussion

495.	A water molecule can from a maximum of ........... Hydrogen bonds.A. 1B. 2C. 3D. 4
Answer» Correct Answer - D

Discussion

496.	The process of cell death involving DNA cleavage in cells is known as- (A) necrosis (B) apoptosis (C) cytokinesis (D) endocytosis
Answer» Correct Option :- (B) apoptosis

496.

The process of cell death involving DNA cleavage in cells is known as- (A) necrosis (B) apoptosis (C) cytokinesis (D) endocytosis

Answer»

Correct Option :- (B) apoptosis

Discussion

497.	In the circuit shown below (on the left) the resistance and the emf source are both variable. The graph of seven readings of the voltmeter and the ammeter (V and I, respectively) for different setting of resistance and the emf, taken at equal intervals of time `Deltat`, are shown (on the right) by the dots connected by the curve EFGH. Consider the interval resistance of the battery to be negligible and the voltmeter and ammeter to be ideal devices. Take `R_(0)=V_(0)//l_(0)`. Then the plot of the resistance as a function of time corresponding to the curve EFGH is given byA. B. C. D.
Answer» Correct Answer - D

Discussion

498.	`2.52 g` of oxalic acid dehydrate was dissolved in 100 ml of water, 10 mL of this solution was diluted to 500 mL. The normality of the final solution and the amount of oxalic acid (mg/mL) in the solution are respectively-A. `0.16 N, 5.04`B. `0.08 N, 3.60`C. `0.04 N, 3.60`D. `0.02 N, 10.08`
Answer» Correct Answer - C Initial Normality `N = (2.52 xx 1000)/(63 xx 100) = 0.4` `{:(underset(\|)(C)OOH),(COOH):}.2H_(2)O` `:.N_(1)V_(1)=N_(2)V_(2)` `0.4 xx 10 = N_(2) xx 500` `N_(2) = (0.4)/(50) = 0.08 N` Then final weight `N = (w xx 100)/(E xx V_(ml))` `0.08 = (w xx 1000)/(63 xx 500)`

Discussion

499.	the number fo electrons required to reduce chromium completely in `Cr_(2) O_(7) ^(2-) ` to `Cr_(3+)` in acidic meddium , isA. 5B. 3C. 6D. 2
Answer» Correct Answer - C `Cr_(2) O_(7) ^(-2) +14H^(+)+6e^(-) to 2Cr^(+3) +7H_(2) O` `therefore ` no of electron = 6

Discussion

500.	Using the Heisenberg uncertainty principle, arrange the following particles in the order of increasing lowest energy possible. (I) an electron in H2 molecule (II) a H atom in a H2 molecule (III) a proton in the carbon nucleus (IV) a H2 molecule within a nanotube (A) (I) < (III) < (II) < (IV) (B) (IV) < (II) < (I) < (III) (C) (II) < (IV) < (III) < (I) (D) (IV) < (I) < (II) < (III)
Answer» Correct option (B) (IV) < (II) < (I) < (III) Explanation: By theory.

500.

Using the Heisenberg uncertainty principle, arrange the following particles in the order of increasing lowest energy possible. (I) an electron in H2 molecule (II) a H atom in a H2 molecule (III) a proton in the carbon nucleus (IV) a H2 molecule within a nanotube (A) (I) < (III) < (II) < (IV) (B) (IV) < (II) < (I) < (III) (C) (II) < (IV) < (III) < (I) (D) (IV) < (I) < (II) < (III)

Answer»

Correct option (B) (IV) < (II) < (I) < (III)

Explanation:

By theory.

Discussion

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