This section includes 7 InterviewSolutions, each offering curated multiple-choice questions to sharpen your Current Affairs knowledge and support exam preparation. Choose a topic below to get started.
| 1. |
At a wedding, the bride usually wears a ___dress. |
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Answer» Correct answer is white |
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| 2. |
The system shown in the figure is in equilibrium. Masses `m_(1) & m_(2)` are 2kg and 8 kg respectively. The compression in right spring is 0.5 m. (Both spring have the same natural length) `[K_(1)=20 N//m, K_(2)=70 N//m]`. Choose the correct option(s) : A. elongation in left spring is 0.5mB. elongation in left spring is 0.25 mC. Tension is string is 45 ND. Just after left spring is cut `m_(1)`, has upward acceleration. |
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Answer» Correct Answer - A::B::C F = 8t `m(dv)/(dt) = 8t` `m underset(0)overset(v)int dv = 8 underset(0) overset(t) int t dt` `mv = 4t^(2)` or `m(dx)/(dt) = 4t^(2) implies dx = (4t^(2))/(m) dt = 2t^(2) dt` `W = intFdx = underset(0)overset(2)int (8t xx 2t^(2)) dt = 64J` |
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| 3. |
A box is failing freely. Inside the box, a particle is projected with some velocity v with respect to the box at an angle `theta` as shown in the figure. A. The path of the particle with respect to an observer sitting in the box will be a straight line.B. The acceleration of particle with respect to the box is zero.C. The path of the particle with respect to an observer sitting in the box will be a parabolaD. The acceleration of particle with respect to ground is g downward |
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Answer» Correct Answer - A::B::C Loss in KE = `(1)/(2) (m_(1)m_(2))/(m_(1)+m_(2))(V_("rel"))^(2)(1-e^(2))` |
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| 4. |
A uniform field is exists in the region directed away from the page. A charged particle, moving in the plane of the page follows a anticlockwise spiral of increasing radius as shown. True explanation is: A. the charge is positive and slowing downB. the charge is negative and slowing downC. the charge is positive and speeding upD. the charge is negative and speeding up |
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Answer» Correct Answer - A `R=(mv)/(qB)` : `vec(F) = q(vec(v)xxvec(B))` |
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| 5. |
A block of mass m is placed on a smooth block of mass `M = m` with the help of a spring as shown in the figure. A velocity `v_(0)` is given to upper block when spring is in natural length. Find maximum compression in the spring. A. `sqrt((m)/(2K)) V_(0)`B. `sqrt((m)/(K)) v_(0)`C. `sqrt((2m)/(K)) v_(0)`D. `2 sqrt((m)/(K)) v_(0)` |
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Answer» Correct Answer - A Compression will be maximum when relative velocity between blocks is zero. Means both have Sam velocity apply conservation of momentum `mv_(0) = (M + m) v` `rArr v = (m)/(M + m) v_(0)` Apply conservation of kinetic energy `(1)/(2) mv^(2) + (1)/(2) Mv^(2) + (1)/(2) kx^(2) = (1)/(2) mv_(0)^(2)` `rArr (m^(3)v_(0)^(2))/((M + m)^(2)) + (m^(2)Mv_(0)^(2))/((M + m)^(2)) + kx^(2) = mv_(0)^(2)` `rArr kx^(2) = (m^(3) + m^(2)M)/((M + m)^(2)) v_(0)^(2) = mv_(0)^(2)` `rArr kx^(2) = v_(0)^(2) [(m)/(p) - (m^(2) (M + m))/((M + m)^(2))]` `= v_(0)^(2) [(M_(m) + m^(2) - m^(2))/(M + m)] = (M_(m)v_(0)^(2))/(M + m)` `rArr x = sqrt((Mm)/(k(M + m))) v_(0)` |
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| 6. |
A uniform field is exists in the region directed away from the page. A charged particle, moving in the plane of the page follows a anticlockwise spiral of increasing radius as shown. True explanation is: A. charge is positive and slowing downB. charge is negative and slowing downC. charge is positive and speeding upD. charge is negative and speeding up |
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Answer» Correct Answer - D `R=(mV)/(qB) , R uarr , v uarr` |
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| 7. |
Water kept in a porcus pot evaporates through the walls of pot. Rate of evaporation is proportional to volume of water. When water is kept in the pot 75% water get evaporated in 16 hrs. A suction mechanism attached in the pot sucks water at the rate which is also proportional to volume of water in the pot. Without evaporations, half of the water kept in pot is sucked in 24 hr The pot is filled with 16 kg of water, with evaporation and suction acting simultaneously , what amount of water (in kg) will be left is pot after one day. |
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Answer» Correct Answer - `0001` Evaporations and reaction has rate similar to first order reaction rate hence `1/t_(1//2)=1/((t_(1//2))_(evoparation))+1/((t_(1//2))_(suction))Rightarrow1/(t_(1//2))=6`hrs `Hence water left =16/2^(4)=1kg` |
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| 8. |
The reaction `_(3)^(7)Li+_(1)^(1)H rightarrow _(4)^(7)Be + _(0)^(1)n` is endothermic. Assuming that Li nuclei is free and at rest. What is the minimum kinetic energy ( in keV) of incident proton so that this reaction occurs? Take Q value of this reaction as -1645 keV. |
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Answer» Correct Answer - `1880 ke V` Energy available `=1/2mu v_(rel)^(2)=Q` value `=1/2xx(7xx1)/(7+1)xxv_(rel)^(2)=Q` value `Rightarrow 1/2xxv_(rel)^(2)=Qxx8/7` `K_(i)=1645xx8/7=1880` ke V |
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| 9. |
In a sample initially there are equal number of atoms of two radioactive isotopes A and B. 3 days later the number of atoms of A is twice that of B. Half life of B is `1.5` days. What is half life of isotope A? (in days) |
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Answer» Correct Answer - `0003` `A B t=0 N_(0) N_(0)` `t_(0)=3 days 2N N` `2N=N_(0)(0.5)t_(0)//tau_(1)` `N=N_(0)(0.5) t_(0)//tau_(2)` `2=(0.5) t_(0)(1/tau_(1)-1/tau_(2))` `Rightarrow 0.5^(-1)=(0.5)(3/tau_(1)-2)` `Rightarrow -1=3/tau_(1)-2 :.tau_(1)=3` days |
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| 10. |
In a slow reaction, heat is being evolved at a rate about 10 m W in a liquid. If the heat were being generated by the decay `of ^(32)` P, a radioactive isotope of phosphorus that has half-life of 14 days and emits only beta-particles with a mean energy of 700KeV, estimate the number `of ^(32)` P atoms in the liquid. Express your answer in form of `Axx10^(15)` and fill A in OMR sheet. Round off A to nearest integer [Take:l n`2=0.7`] |
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Answer» Correct Answer - `0154` `P=700xx10^(3)xx1.6xx10^(-19)xx(dN)/(dt)=10xx10^(-3)` `(dN)/(dt)=10^(-2)/10^(-14)xx1/(7xx16)=10^(12)/11.2=lambda N_(0)` `lambda=(l n2)/(14xx86400) Rightarrow N_(0)=(14xx86400xx10^(12))/(11.2l n2)=154xx10^(15)` |
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| 11. |
A capstan is a rotating drum (cylinder) over which a rope or cord slides in order to increase the tension due to friction. If the difference in tension between the two ends of the rope is 500 N and the capstan has a diameter of 10 cm and rotates with angular velocity `10 rad//s`. Capstan is made of iron and has mass 5 kg, specific heat `1000 J//kg` K. At what rate does temperature rise? Assume that the temperature in the capstan is uniform and all the thermal energy generated flows into it. Express your answer as `xxx10^(-4)^(@)C` Fill up value of x. |
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Answer» Correct Answer - `0500` `P=tauomega=(500xx5xx10^(-2))xx10` `P=ms(dT)/(dt)` `(dT)/(dt)=500xx5xx10^(-2)xx10=5xx10^(-2)` `x xx10^(-4)=5xx10^(-2)=500` |
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| 12. |
Interference frings of yellow light of wavelength 6000 A are formed by Billet split lenses. The distance from source to lens is 24 cms. The focal length of lens is 15 cm The lens halves are separated by `0.06` mm. The distance of source to screen is 200 cms. Calculate the fringe width (in mm) Round off the answer to nearest integer. |
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Answer» Correct Answer - `5 mm` `u=-24` `v=(uf)/(u+f)` `v=(-24xx15)/(-24+15)=(124xx15)/8=40` `Rightarrow D=200-24-40=136` cm `(R_(1))/(-0.03)=v/u=40/(-24)` `R_(1)=40/24xx0.03=0.05` mm `Rightarrow d=0.05xx2+0.06=0.16xx10^(-3)`m `beta=(lamdaD)/d=(6000xx10^(-10)xx136xx10^(-2))/(0.16xx10^(-3))` `beta=51xx10^(-4)m=5.1` mm |
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| 13. |
In a modified YDSE the sources S of wavelength 5000 A oscillates about axis of setup according to the equation `y=0.5 sin(pi/6)t` where y is in millimeter and t in second. At what time ti will the intensity at P, a point exactly in front of slit `S_(1)` be maximum for the first time? |
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Answer» Correct Answer - `0001` The path difference at point P, `Deltax=(SS_(2)-SS_(1))+(S_(2)P-S_(1)P)` `=(dy)/(D_(1))+(d(d//2))/D_(2)` For constructive interference, `Deltax=(dy)/D_(1)+d^(2)/(2D_(2))=nlambda` `((10^(-3))(0.5sinpit)xx10^(-3))/1+(10^(-3))^(2)/(2xx2)=nlambda` `(0.5sin(pi/6)t)xx10^(-6)+0.25xx10^(-6)` `=(5000xx10^(-10)) n=0.5xx10^(-6)`n `sin(pi/6)t=(0.5n-0.25)/0.5` For the minimum value of t n=1 `sin(pi/6)t=1/2 Rightarrow (pi/6)t=pi/6 or t=1 sec` |
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| 14. |
The spatial distribution of the electric field due to charges `(A,B)` is shown in figure. Which of the following statements is correct? A. `A` is `+ve` and `B-ve` and `|A|gt|B|`B. `A` is `-ve` and `B+ve, |A|=|B|`C. Both are `+ve` but `AgtB`D. Both are `-ve` but `AgtB` |
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Answer» Correct Answer - a Electric lines of force usually start (i.e., diverge out) from positive charge and end(i.e., converge) on negative charge or extendes to infinity. Thus, `A` is positive charge and `B` is negative charge. Also density of lines at `A` is more than that at `B`, i.e., `|A|gt|B|`. |
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| 15. |
If the sum of two unit vectors is a unit vector, then magnitude of difference is-A. `sqrt(2)`B. `sqrt(3)`C. `(1)/sqrt(2)`D. `sqrt(5)` |
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Answer» Correct Answer - B `|hatn_(2)+hatn_(2)|=1` `sqrt(1^(2)+1^(2)+2 cos theta)=12+2 cos theta=1` `rArr costheta=(-1)/(2)rArrtheta=120^(@)` `|hatn_(1)-hatn_(2)|=sqrt(1^(2)+1^(2)-2xx1xxa cos 120^(@))=sqrt(3)` |
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| 16. |
Which of the following sets of concurrent force may be in equilibrium?A. `F_1= 3N, F_2= 5N, F_3= 1N`B. `F_1 = 3N, F_2 = 5N, F_3= 9N`C. `F_1 = 3N, F_2= 5N, F_3= 6N`D. `F_1= 3N, F_2= 5N, F_3= 15N` |
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Answer» Correct Answer - 3 `F_1 - F_2 le |vecF_1 + vec F_2| le F_1 + F_2 ` To produce zero resultant `F_3` must lie between `F_1 - F_2 le F_3 le F_1 + F_2` |
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| 17. |
What is the torque of the force `vecF=(2hati+3hatj+4hatk)N` acting at the point `vecr=(2hati+3hatj+4hatk)m` about the origin? (Note: Tortue, `vectau=vecrxxvecF`) |
| Answer» `"Torque" vectau=vecrxxvecF=|{:(hati,hatj,hatk),(2,3,4),(2,3,4):}|=hati(12-12)-hatj(8-8)+hatk|6-6|=0hati-0hati-0hatj+ohatk=vec0` | |
| 18. |
The magnitudes of vectors `vecA.vecB` and `vecC` are respectively 12,5 and 13 unira and `vecA+vecB=vecC`, then the angle between `vecA` and `vecB` is : |
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Answer» Correct Answer - C `vecA+vecB=vecC` as `vecA^(2)+B^(2)=C^(2)` thus angle between A and B is `90^(@)` |
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| 19. |
Potential energy of a particle of mass `m`, depends on distance `y` from line `AB` according to given relation `U = (K)/(sqrt(y^(2) + a^(2))`, where `K` is a positive constant. A particle of mass `m` is projected from `y = sqrt(3)` towards line `AB`. (perpendicular to it) then minimum velocity so that it connot return to its initial point is `sqrt((K)/(aNm))`, calculate `N`. |
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Answer» Correct Answer - 1 `U_(i) + k_(i) = U_(f) + k_(f)` `(K)/(2a) + (1)/(2)mV_(0)^(2) = (K)/(a)` `(mV_(0)^(2))/(2) = (K)/(2a) rArr v_(0) = sqrt((K)/(am))` |
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| 20. |
All the pulleys are ideal, string is massless then rate of work done by gravity at the given instant is `(-x xx 10^(2))W` then calculate `x` : |
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Answer» Correct Answer - 5 `v_(p) = (-5 + 15)/(2) = 5m//s` `P = overset(vec)(F).overset(vec)(V) = -mgV` `= -100 xx 5 = -500 W` |
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| 21. |
If the energy ( E) ,velocity (v) and force (F) be taken as fundamental quantities,then the dimension of mass will beA. `Ev^2`B. `Ev^-2`C. `Fv^-1`D. `Fv^-2` |
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Answer» Correct Answer - B Let `mpropE^xv^yF^z` By substituting the following dimensions: `[E]=[ML^2T^-2]`,`[v]=[LT^-1]`,`[F]=[MLT^-2]` and by equating the both sides |
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| 22. |
The dimensions of `a/b` in the equation `P=(a-t^(2))/(bx)` where `P` is pressure, `x` is distance and `t` is time areA. `[M^(2)LT^(-3)]`B. `[MT^(-2)]`C. `[LT^(-3)]`D. `[ML^(3)T^(-1)]` |
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Answer» Correct Answer - B `[a]=T^(2)` `[X]=L` `[P]=ML^(-1)T^(-2)=T^(2)/[b]L` `[b]=T^(2)/ML^(-1)T^(-2)L=M^(-1)T^(4)` `[a]/[b]=T^(2)/M^(-1)T^(4)=MT^(-2)` |
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| 23. |
If pressure P, velocity V and time T are taken as fundamental physical quantities, the dimensional formula of force ifA. `PV^2T^2`B. `P^-1V^2T^-2`C. `PVT^2`D. `P^-1VT^2` |
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Answer» Correct Answer - A Let `Fpropp^xV^yT^z` by subsituting the following dimensions: `[P]=[ML^-1T^-2][V]=[LT^-1]`,`[T]=[T]` and comparing the dimension of both sides `x=1`,`y=2`,`z=2`, so `F=PV^2T^2` |
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| 24. |
If the mass time and work are taken as fundamental physical quantities then dimensional formula of lengthA. `[m^(1/2)T^(1)W^(-1/2)]`B. `[M^(-1/2)T^(1)W^(1/2)]`C. `[M^(-1)T^(2)W]`D. none of these |
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Answer» Correct Answer - B `L=M^(a)T^(b)(ML^(2)T^(-2))^(c)` `a+c=0` ` 2c=1` `c=1/2 a= (-1)/2` ` b=2c=0 rArr b=1` |
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| 25. |
Consider a screw gauge without any zero error. What will be the final reading corresponding to the final state as shown? It is given that the circular head translates P msd in N rotations One msd is equal to 1 mm. A. `(P/N)(2+45/100)`mmB. `(N/P)(2+45/N)`mmC. `P(2/N+45/100)`mmD. `[2+45/100xxP/N]` mm |
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Answer» Correct Answer - D Nrotations rightarrow P msd ` 1 rotation (100 divisions) rightarrow P/N msd` `45 division rightarrow 1/100xxP/Nxx45 msd` (mm) The reading shows two msd and 45 on the circular theremore reading `= 2mm + 1/100xxP/Nxx45` (mm) |
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| 26. |
The diagram shows part of the vernier scale on a pair of calipers Which reading is correct A. 2.74cmB. 3.10cmC. 3.26cmD. 3.64cm |
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Answer» Correct Answer - A VS `=4 MS = 2.7` `MS +VS xx LC` ` 2.7 + 0.04= 2.74` |
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| 27. |
A unit positive point charge of mass m is projected with a velocity v inside the tunnel as shown. The tunnel has been made inside a uniformly charged nonconducting sphere. The minimum velocity with which the point charge should be projected such it can it reach the opposite end of the tunnel is equal to A. `[rhoR^2//4mepsilon_0]^1/2`B. `[rhoR^2//24mepsilon_0]^1/2`C. `[rhoR^2//6mepsilon_0]^1/2`D. zero because the initial and the final points are at same potential |
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Answer» Correct Answer - A If we throw the charged particle just right of the `v` center of the tunnel, the particle will cross the tunnel. Hence applying conservation of `ME` between start point and center of tunnel, `DeltaK+DeltaU=0` or `(0+(1)/(2)mv^(2))+q(V_(f)-V_(i))=0` or `V_(f)=(V_(s))/(2)(3-(r^(2))/(R^(2)))=(pR^(2))/(6epsilon_(0))(3-(r^(2))/(R^(2)))` Hence `r=(R)/(2)` `V_(f)=(pR^(2))/(6epsilon_(0))*(3-(R^(2))/(4R^(2)))=(11pR^(2))/(24epsilon_(0))` `V_(i)=((pR^(2))/(3epsilon_90))` `(1)/(2)mv^(2)=1[(11pR^(2))/(24epsilon_(0))-(pR^(2))/(3epsilon_90)]=(pR^(2))/(3epsilon_(0))[(11)/(24)-(1)/(3)]` `=(pR^(2))/(8epsilon_(0))` or `V=((pR^(2))/(4mepsilon_(0)))^(1//2)` Hence, velocity should be slightly greater than `V`. |
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| 28. |
Two insulting plates are both uniformly charged in such a way that the potential difference between them is `V_2-V_1=20V`. (i.e., plate 2 is at a higher potential). The plates are separated by `d=0.1m` and can be treated as infinity large. An electron is released from rest on the inner surface of plate 1. What is its speed when it hits plate 2? (`e=1.6xx10^-19C`, `m_e=9.11xx10^-31kg`) A. `7.02 xx 10^(12)m//s`B. `1.87 xx10^(6)m//s`C. `32 xx 10^(-19) m//s`D. `2.65 xx 10^(6)m//s` |
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Answer» Correct Answer - D `(1)/(2) mv^(2)=q Deltav` `V^(2) = (2q Delta V)/(m) =(2xx1.6 xx 10^(-19)xx20)/(9.11 xx 10^(-31))` `V^(2)=7.02 xx 10^(12)` `V=2.65 xx 10^(6)m//s` |
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| 29. |
A few electric field lines for a system of two charges `Q_(1)` and `Q_(2)` fixed at two different points on the `x`-axis are shown in the figure. These lines suggest that (i) `|Q_(1) | gt |Q_(2)|` (ii) `|Q_(1)| lt |Q_(2)|` (iii) At a finite distance to the left of `Q_(1)` the electric field is zero (iv) At a finite distance to the right of `Q_(2)` the electric field is zero A. `|Q_(1)| gt |Q_(2)|`B. `|Q_(1)| lt |Q_(2)|`C. at a finite distance to the left of `Q(1)` the electric field is zeroD. at a finite distance distance to the right of `Q_(2)` the electric field is zero. |
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Answer» Correct Answer - A::D |
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| 30. |
What are isotopes? |
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Answer» The nuclei which have the same atomic number but different mass numbers. |
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| 31. |
What is ‘diffraction of light’? Explain its two types |
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Answer» The bending of light near the edge of an obstacle or slit and spreading into the region of geometrical shadow is called diffraction of light. Two types of diffraction are: a. Fresnel diffraction: Diffraction pattern in which source of light and screen are kept at finite distance from the slit is called fresnel diffraction. eg: Diffraction of straight edge, small opaque disc, narrow rectangular slit, etc. b. Fraunhofer diffraction: Diffraction pattern in which, the source of light and the screen are effectively at infinite distances from the diffracting system is called Fraunhoffer diffraction. In this diffraction pattern convex lens is used. eg: Diffraction due to single slit, double slit, etc. |
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| 32. |
Distinguish between ‘paramagnetic’ and ‘ferromagnetic’ substances. |
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| 33. |
Mention one need for modulation. |
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Answer» 1. To reduce the size of the antenna 2. Effective power radiated by the antenna 3. Mixing up of signals from different transmitters |
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| 34. |
Write any two limitations of Ohm’s law. |
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Answer» 1. It is not applicable for metallic conductor at very low and very high temperatures. 2. It is not applicable for semiconductors, super conductors, triodes. |
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| 35. |
Column I may match with more than one conditions of column II. `{:("Column-I","Column-II"),((A)Ph-CH=CH-Ph,(p)"Ozonolysis followed by reaction with" `NH_2OH` "leads to more than one oxime prouduct"),((B)(CH_3)_2C=CH-undersetunderset(CH_3)(|)CH-Cl,(q)"Can exhibit geometrical isomers"),(( C)CH_2=CH-CH=CH_2,(r)"Compounds with this structure formula can be separated into different fractions upon fractional distillation"),((D)OHC-undersetunderset(OH)(|)(CH)CH-undersetunderset(OH)(|)(CH)-CHO,(s)"is capable of showing steroisomerism"),(,(t)"On hydrogenation it gives more than one product"):}` |
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Answer» Correct Answer - A-p,q,r,s ; B-p,s,t ; C-p ; D-r,s,t (A)`Ph-CH=CH-Phoverset(O_3//Zn)toPh-CHOoverset(NH_2OH)tounderset(("syn/anti"))(Ph-CH=N-OH)` Ph-CH=CH-Ph shows geometrical isomers and can be sepearated by fractional distillation. `Ph-CH=CH-PHh+H_2toPh-CH_2-CH_2-Ph`(one product) (B)`(CH_3)_2C=CH-undersetunderset(CH_3)(|)CH-Cloverset((O_3//Zn))toCH_3-undersetunderset(O)(||)C-CH_3+CH_3-undersetunderset(Cl)(|)CH-CHOoverset(NH_2OH)to`more than one oxime `(CH_3)_2C=CH-undersetunderset(CH_3)(|)CH-Cl+H_2to(CH_3)_2CH-CH_2-undersetunderset(CH_3)(|)overset(**)CH-Cl`(more than one product) (D)`OHC-undersetunderset(OH)(|)overset(**)CH-undersetunderset(OH)(|)overset(**)CH-CHO+H_2toundersetunderset(OH)(|)CH_2-undersetunderset(OH)(|)overset(**)CH-undersetunderset(OH)(|)overset(**)CH-undersetunderset(OH)(|)CH_2`(more than one product) |
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| 36. |
Match the following : (More than one option in column-II may match with single option in column I |
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Answer» Correct Answer - A-p,q,r,s ; B-r ; C-p,r ;D-q a protic acid gives positive test with Na metal a carboxylic functional group gives positive test with `NaHCO_3` a carbonyl group yellow ppt. with 2,4-DNP an alcohol gives positive test with Lucas Reagent. |
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| 37. |
How many alkenes can be hydrogenated to give an alkane with molecular formula `C_4H_10` which give three monochloro derivative on monochlorination. |
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Answer» Correct Answer - 3 The alkane is n-Butane `CH_3-CH_2-CH_2-CH_3` and total 3 alkenes can be hydrogenated, `CH_3-CH_2-CH=CH_2 " " underset((" cis & trans"))(CH_3-CH=CH-CH_3)` |
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| 38. |
Work in groups and write a paragraph on “Laughter is the best medicine.” |
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Answer» Laughter is the best Medicine A good laugh heals a lot of hurts. It is a mighty good thing that sets everything straight. It is a powerful antidote to stress, pain and conflict. Nothing works faster to bring your mind and body back into balance than a good laugh. Humour lightens your burden, inspires hope, connects you to others and keeps you grounded, focussed and alert. A good hearty laugh relieves physical tension and stress leaving your muscles relaxed for up to 45 minutes after. It improves the function of blood and increases blood flow, which can help protect you against heart attack and other vascular diseases. Shared laughter is one of the most effective tool. When laughter is shared, it binds people together and increases happiness and intimacy. Therefore, one should always maintain positive attitude in life and keep laughing to stay healthy and happy. |
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| 39. |
The elevated activity of amylase in the blood is indication of:A. hepatitisB. muscle dystrophyC. acute pancreatitisD. myocardial infarctionE. stomach disease |
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Answer» Correct option is C. acute pancreatitis |
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| 40. |
Put the words in the correct order to make sentences. Use a different colour to write the adverbs.1. Was / somebody / there / nearby / standing.2. Came / Anand / to / school / early.3. Softly / Murali / speaks4. Beautifully / the house / have / they / decorated5. English / classes / during / always / we / English / speak. |
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Answer» 1. There was somebody standing nearby. 2. Anand came to school early. 3. Murali speaks softly. 4. They have decorated the house beautifully. 5. We always speak English during English classes. |
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| 41. |
Look at the picture and fill in the blanks with suitable words.1. The car was moving too _______2. The lift is moving _______3. Joanna did her classwork _____4. Keerthi Vasan arrived _____5. Paul _______ plays cricket with his friends.6. Suguna _______ helps her mother at home. |
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Answer» 1. fast 2. slowly 3. neatly 4. late 5. always 6. often |
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| 42. |
Just then, the richest farmer in the village pushed his way to the _______ of the group. (a) back (b) middle (c) side(d) front |
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Answer» Correct answer is (d) front |
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| 43. |
It poured and poured and only those of us who have seen the _______ will know what that means. (a) rains (b) showers (c) monsoons(d) cyclones |
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Answer» (c) monsoons |
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| 44. |
Here was water to be had, and so close to his . (a) field (b) land (c) holding (d) crop |
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Answer» Correct answer is (c) holding |
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| 45. |
The dry earth soaked up the moisture, as a hungry _______ laps up milk. (a) cat (b) lion (c) puppy (d) monkey |
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Answer» Correct answer is (c) puppy |
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| 46. |
Name one reagent or one operation to distinguish between: a. `Be(OH)_(2)` and `Ca(OH)_(2)` b. `BeSO_(4)` and `SrSO_(4)` c. `K_(2)CO_(3)` and `KHCO_(3)` |
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Answer» a. `Be(OH)_(2)` dissolves in alkali, but `Ca(OH)_(2)` does not. b.`BeSO_(4)` is soluble in water while `SrSO_(4)` is not. c. `K_(2)CO_(3)` does not decompose on heating, whereas `KHCO_(3)` decomposes on heating to give `CO_(2)`, which turns limewater milky. |
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| 47. |
What are the various factors due to which the ionisation enthalpy of the main group elements tends to decrease down a group? |
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Answer» Within the main group elements, the `IE` decreases regularly as we move down the group due to the following two factors: a. Atomic size.On moving down the group, the atomic size increases gradually due to the addition of one new pricipal energy shell at each succeeding element. As a result, the distance of the valence electrons from teh nucleus increses. Consequently, the force of attraction of the nucleus for the valence electrons decreases and hence the ionisation enthalpy decreases. Screening effect. With the addition of new shells, the number of inner electron shells which shield the valence electrons increases. In other words, the shielding effect or the screening effect increases. As the talence electrons further decreases and hence the `IE` decreases. |
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| 48. |
Why sodium cannot be prepared by electrolysis of its aqueous solution? |
| Answer» The reduction potential of sodium,`Na`,is much lower than that of `H_(2)O`, therefore on electrolysis, water gets reduced to liberate `H_(2)` in preference of `Na^(o+)` ion. Hence, sodium cannot be prepared by electrolysis of its aqueous solution. | |
| 49. |
The first ionisation enthalpy of group `13` elements are : Explain this deviation from the general trend. |
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Answer» On moving down the group `13` from `B` to `Al`,`IE` decreases as expected due to an increase in atomic size and screening effect which outweigh the effect of increased nuclear charge. However, `IE_(1)` of `Ga` is only slighly higher `(2 kJ mol^(-1))` than that of `Al` while that of `TI` is much higher than those of `Al`,`Ga` and `In`. These deviations are due to : `Al` follows immediately after `s`-block elements and `TI` after d-and f-block elements. These extra `d-` and `f-` electrons do not shield (or screen) the outer shell electrons from the nucleus very effectively. As a result, the valence electrons remain more tightly held by the nucleus and hence larger amount of energy is needed for their removal. This explains why `Ga` has higher `IE` than `Al`. Further on moving down the group from `Ga` to In , the increased shielding effect (due to the presence of additional `4d`-electrons) outweighs the effect of increased nuclear charge `(49-31=18 units)` and hence the `IE_(1)` of In is lower than that of `Ga`. Thereafter, the effect of increased nuclear charge `(81-49=32 units)` outweighs the shielding effect due to the presence of additional `4f` and `5d` electrons and hence the `IE_(1)` of `TI` is higher than that of `In`. |
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| 50. |
Which of the following is correct for various reversible reaction ?A. In case of liq `hArr` gas, systems the equilibrium is attained between both the phases, whether the system is closed or open.B. Boiling point of a liquid is independent of the altitude of the placed of experiment.C. The value of equilibrium constant is independent of concentration terms.D. It a reversible reaction has high value of its equilibrium constant, the reaction quickly attains the equilibrium. |
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Answer» Correct Answer - C (A) In open system, said equilibrium is not attained. (B) Atomspheric pressure changes with altitude. (C ) `K_(eq)` is the function of temperature only. (D) `K_(eq)` does not give any information about the rate at which the equilibrium is reached. |
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