This section includes InterviewSolutions, each offering curated multiple-choice questions to sharpen your knowledge and support exam preparation. Choose a topic below to get started.
| 12501. |
order of magnitude of acceleration due to gravity is ........(g=9.8m/s²) |
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Answer» order of magnitude is given by the power of 10 in any quantity acceleration due to gravity 9.8 m/s2 so order is 10 per power 0 |
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| 12502. |
Two closed organ pipes give `10` beats between the fundamental when sounded together. If the length of the shorter pipe is `1m` then the length of the longer pipe will be (speed of sound in air is `340ms^(-1)`)A. `2.87 m`B. `0.87 m`C. `1.13 m`D. `2.13 m` |
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Answer» Correct Answer - C `(v)/(4)((1)/(l_(1))-(1)/(l_(2)))=10`, Here `l_(1)=1m`, `v=340ms^(-1)` |
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| 12503. |
The oscillations represented by curve 1 in the graph are expressed by equation`x=Asinomegat.` The equation for the oscillation represented by curve `2` is expressed asA. `x=2Asin(omegat-(pi)/(2))`B. `x=2Asin(omegat+(pi)/(2))`C. `x=-2Asin(omegat-(pi)/(2))`D. `x=Asin(omegat-(pi)/(2))` |
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Answer» Correct Answer - A Oscillation 1 has amplitude `=A,2` has amplitude `=2A` But (2) lags by phase `(pi)/(2)`. Also, putting `t=0` in option `A.X=-2A` |
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| 12504. |
The oscillations represented by curve 1 in the graph are expressed by equation`x=Asinomegat.` The equation for the oscillation represented by curve `2` is expressed asA. `x=2A sin (omegat-pi//2)`B. `x=2Asin(omegat+pi//2)`C. `x=-2Asin(omegat-pi//2)`D. `x=A sin (omegat-pi//2)` |
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Answer» Correct Answer - A Oscillations represented by curve 2 lags in phase by `pi//2` and the periods are same. Amplitudes of curve 2 is double that of 1. Hence (1). |
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| 12505. |
Radioactive nuclei emit `beta^-1` particles. Electrons exist inside the nucleus. |
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Answer» Correct Answer - C For a `beta^(-)` decay. `_(0)^(1)to _(1)^(1)p + _(-1)^(0)e` `therefore` Electrons do not exist inside the nucleus. |
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| 12506. |
Assertion `:` To hear distinct be beats, difference in frequencies of two sources should be less than 10. Reason `:` More the number of beats `//` sec., more is the confusion.A. Both Assertion and Reason are correct, Reason is the coreect expianation of AssertionB. Bioth Assertion and Reason are Correct but Reason is not the correct expalnation of AssertionC. Assetion is correct and Reason is incorrectD. Assertion is incorrect and Reason is correct |
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Answer» Correct Answer - A |
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| 12507. |
It was raining when I ______ the house this morning. A) leave B) leaving C) left D) have left |
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Answer» Correct option is C) left |
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| 12508. |
How would you explain the reluctance of the soldier to leave the shelter of the doctor’s home even when he knew he could not stay there without risk to the doctor and himself? |
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Answer» When the American war prisoner came to consciousness and realized that he was saved by a Japanese family, he feared that he will be soon handed over to the army. However, as he noticed the amount of concern and care given to him by the family, he understood that he was in safe hands. He knew that although he was a threat to the doctor’s family, his own life might be saved there. Burdened with gratitude towards the family, he ultimately decidesto comply withwhat the doctor planned for him – the escape. |
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| 12509. |
A particle of mass `5kg` moving in the `X-Y` plane has its potential energy given by `U=(-7x+24y) J`. The particle is initially at origin and has a velocity `vec(U)=(14.4+4.2J)m//s`A. The particle has speed `20 m//s` at `t=4` secB. The particle has an acceleration `25 m//s^(2)`C. The acceleration of particle is normal to its intial velocity.D. None of the above each correct |
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Answer» Correct Answer - C Force acting on the particle `vec(F)=-(delU)/(delx)hat(i)-(delU)/(dely)hat(j)=7hat(i)-24hat(j)` Accerleration of particle `=(F)/(m)=sqrt(7^(2)+24^(2))/(5)=(25)/(5)=5ms^(-2)` velocity of particle `vec(v)=vec(u)+vec(a)t=(14.4hat(i)+4.2hat(j))+(1.4hat(i)-4.8hat(j))t` Speed at `4` sec`=sqrt((14.4+5.6)^(2)+(19.2-4.2)^(2))` `=25ms^(-1)` Now `vec(a).vec(u)=(1.4hat(i)-4.8hat(j)).(14.4hat(i)+4.2hat(j))=20.16-20.16=0rArr` The acceleration of particle is normal to its initial velocity. |
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| 12510. |
The potential energy for a force filed `vecF` is given by `U(x,y)=cos(x+y)`. The force acting on a particle at position given by coordinates `(0, pi//4)` isA. `-(1)/(sqrt2)(hat(i)+hat(j))`B. `(1)/(sqrt2)(hat(i)+hat(j))`C. `((1)/(2)hat(i)+(sqrt3)/(2)hat(j))`D. `((1)/(2)hat(i)-(sqrt3)/(2)hat(j))` |
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Answer» Correct Answer - B `F_(x)=-(delU)/(delx)=sin(x+y)` and `F_(y)=-(delU)/(dely)=sin(x+y)` so `F_(x)=[sin(x+y)]_(((0,pi/4)))=(1)/(sqrt2)` `F_(y)=[sin(x+y)]_(((0,pi/4)))=(1)/(sqrt2)` `vec(F)=(1)/(sqrt2)[hat(i)+hat(j)]` |
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| 12511. |
In a process, the number of cycles to failure decreases exponentially with an increase in load. At a load of 80 units, it takes 100 cycles for failure. When the load is halved, it takes 10000 cycles for failure. The load for which the failure will happen in 5000 cycles is . (A) 40.00 (B) 43.25(C) 60.01 (D) 92.02 |
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Answer» Correct option is (B) 43.25 From the data given we assume load = exponent/log(cycles) 80 = x/log(10000) ⇒ x = 160 40 = x/log(10000) ⇒ x = 160 load = 160/log5000 = 43.25 |
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| 12512. |
The potential energy of a force filed `vec(F)` is given by `U(x,y)=sin(x+y)`. The force acting on the particle of mass m at `(0,(pi)/4)` isA. 1B. `sqrt2`C. `(1)/(sqrt2)`D. 0 |
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Answer» Correct Answer - A `vec(F) = DeltaU = - cos (x + y) hat(i) - cos (x + y) hat(j)` `vec(F) |_(((0, pi //4))) = 1` |
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| 12513. |
In the given figurethe wedge is fixed, pulley is frictionless and string is light. Surface AB is frictionless whereas AC is rough. If the block of mass 3m. Slides down. With conttant velocity, then the coefficient of friction between surface AC and the block is A. `(1)/(3)`B. `(2)/(3)`C. `(1)/(2)`D. `(4)/(3)` |
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Answer» Correct Answer - C If a moves down the incline by 1m, B shall move up `hv(1)/(2) m` If the speed of B is v then the speed of A will be 2v. From conservation of energy, Gain in KE = loss in PE `(1)/(2) m_(A) (2v)^(2) +(1)/(2) m_(B)v^(2)=m_(A) g xx (3)/(5)-m_(B)g xx (1)/(2)` soving we get `v = (1)/(2) sqrt((g)/(3))` |
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| 12514. |
Two blocks are connected by a spring and given velocity `v_(1)` and `v_(2)` as shown in figure when spring is unstrected Statement-1: In centre of mass frame, both the blocks come to rest simultaneously Statement-2: Momentum of a system in centre of mass frame is always zero.A. Statement-1 is true, statement-2 is true and statement-2 is correct explanation for statement-1.B. Statement-1 is true, statement-2 is true and statement-2 is NOT the correct explanation for statement-1.C. Statement-1 is true, statement-2 is false.D. Statement -1 is false, statement -2 is true. |
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Answer» Correct Answer - A Momentum of a system is zero in C-frame so if 2 block system is true and one comes to rest w.r.t.c.m. other will also be at rest to make momentum zero |
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| 12515. |
Two students were given a physics problem for finding maximum extension of spring if blocks are imparted velocities `v_(1)` & `V_(2)` when spring is unstrected. By Students A: `1/2m(v_(1)+v_(2))^(2)=1/2kx^(2)` By Students B: `1/2mv_(1)^(2)+1/2mv_(2)^(2)=1/2kx^(2)`A. Student A is correct, Student B is wrongB. Student B is correct, Student A is wrongC. both are correctD. both are wrong |
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Answer» Correct Answer - D Velocity of centre of mass is non zero `1/2(m/2)(v_(1)+v_(2))^(2)=1/2kx^(2)max` |
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| 12516. |
The amplitude of a particle executing SHM is 4 cm. At the mean position, the speed of the particle is 16 `cms^(-1)`. The distance of the particle from the mean position at which the speed of the particle becomes `8 sqrt(3) cms^(-1)` will beA. `2sqrt3` cmB. `sqrt3` cmC. 1 cmD. 2 cm |
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Answer» Correct Answer - D At mean position velocity is maximum i.e., `v_"max" = omega a rArr omega = v_"max"/a =16/4=4` `v=omega sqrt(a^2-y^2) rArr 8sqrt3 = 4sqrt(4^2-y^2)` `rArr 192=16(16-y^2) rArr 12=16-y^2 rArr y=2 cm ` |
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| 12517. |
The superposing waves are represented by the following equations : `y_(1) 5 sin 2 pi (10 t-0.1 x ), y_(2)=10 sin 2 pi (20t -0.2 x)` Ratio of intensities `(I_("max"))/(I_("min")) ` will beA. 1B. 9C. 4D. 16 |
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Answer» Correct Answer - B `a_1=5 , a_2=10` `rArr l_"max"/l_"min" =((a_1 + a_2 )^2)/((a_1-a_2)^2) =((5+10)/(5-10))^2 = 9/1` |
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| 12518. |
The ratio of the speed of sound in nitrogen gas to that in helium gas, at 300K isA. `sqrt((2//7))`B. `sqrt((1//7))`C. `(sqrt(3))//5`D. `(sqrt(6))//5` |
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Answer» Correct Answer - B `((C)N_(2))/((C)He) = sqrt((M_(He))/(M_(N_(2)))) = sqrt((4)/(28)) = sqrt((1)/(7))`. |
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| 12519. |
If the speed of light c, acceleration due to gravity (g) and pressure (p) are taken as the fundamental quantities then the dimension of gravitational constant isA. `c^2g^0p^-2`B. `c^0g^2p^-1`C. `cg^3p^-2`D. `c^-1g^0p^-1` |
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Answer» Correct Answer - B Let `[G]propc^xg^yp^z` by substituting the following dimensions: `[G]=[M^-1L^3T^-2]`,`[c]=[LT^-1]`,`[g]=[LT^-2]` and by conparing the power of both sides we cen get `x=0`,`y=2`,`z=-1` `[G]propc^0g^2p^-1` |
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| 12520. |
The `(x, y)` co-ordinates of the corners of a square plate are `(0, 0)`, `(L, L)` and `(0, L)`. The edges of the plate are clamped and transverse standing waves are set up in it. If `u(x, y)` denotes the displacement of the plate at the point `(x, y)` at some instant of time, the possible expression `(s)` for `u` is (are) `(a = positive constant)`A. `a cos(pi r//2 L) cos(pi y//2 L)`B. `a sin(pi x//L) sin(pi y//L)`C. `a sin (pi x//L) sin (2 pi y//L)`D. `a cos(2 pi x//L) sin (pi y//L)` |
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Answer» Correct Answer - B::C Due to the champing of the square plate at the edges, its displacements along the `x-` and y-axes will individually be zero at the edges. Only the choices (b) and ( c) predict these displacements correctly. This is because `sin 0 =0`. |
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| 12521. |
A transverse sinusoidal wave of amplitude `a`, wavelength `lambda` and frequency `f` is travelling on a stretched string. The maximum speed of any point in the string is `v//10`, where `v` is the speed of propagation of the wave. If `a = 10^(-3)m` and `v = 10ms^(-1)`, then `lambda` and `f` are given byA. `lamda = 2 pi xx 10^-2 m`B. `lamda = 10^-3 m`C. `f = (10^3)/(2 pi) Hz`D. `f = 10^4 Hz` |
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Answer» Correct Answer - A::C For a transverse sinusoidal wave travelling on a string the maximum velocity is `a omega`. Also, the maximum velocity is `(v)/(10)=(10)/(10) =1m//s` `:. a omega =1 rArr 10^(-3) xx 2 pi f=1` `rArr f =(1)/(2pi xx 10^(-3)) =(10^(3))/(2 pi) Hz` The velocity `v = f lamda` `:. lamda = (v)/(f) =(10)/(10^(3)//2pi) = 2pi xx 10^(-2) m`. |
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| 12522. |
Ionic mobility of which of the following 2nd group metal ions is maximum in aqueous solution? (1) Mg2+(2) Ca2+(3) Sr2+(4) Be2+ |
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Answer» Correct option is (3) Sr2+ Smaller the size of ion greater is it's hydration & greater is it's hydrated radii & smaller is ionic mobility. So order of ionic mobility: Be2+ < Mg2+< Ca2+ < Sr2+ < Ba2+ |
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| 12523. |
i am preparing for aiims plz anybody put mock tests for me to get practice |
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Answer» You can get all question and answers related to AIIMS with below link here |
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| 12524. |
If λ is the wavelength of sound wave in air column in the first wavelength in the 3rd mode for closed pipe of length l will be A) 4l/3b) 4l/7c) 4l/5D) 2l/3 |
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Answer» Frequency of closed orgen pipe = \(\frac{(2n-1)v}{4}\) f = \(\frac{(2n-1)v}{4}\) \(\frac{v}{\lambda}=\frac{(2n-1)v}{4l}\) \(\frac{4l}{5}=\lambda\) |
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| 12525. |
Explain Huygen's Principle. |
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Answer» Huygen's Principle: Huygen's principle is based on the following assumption : (i) Each point on the given or primary wave front acts as a source of secondary wavelets, sending out disturbance is all directions in a similar manner as the original source of light source. (ii) The new position of the wave front at any instant (called secondary wave front) is then envelope of the secondary wavelets at that instant. |
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| 12526. |
A metallic sheet is inserted between plates parallel to the plates of a parallel plate capacitor. The capacitance of the capacitor |
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Answer» A metallic sheet is inserted between plate parallel to the plate of a parallel plate capacitor. the capacitance of the capacitor will be increase. |
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| 12527. |
A particle of mass m and charge q is released from rest in a uniform electric field E. |
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Answer» F = qE , a= qE/m , u =0, v2= 2ax mv = √2qExm |
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| 12528. |
A lens is a piece of a transparent substance,bounded by two surfaces, out of which one must be curved. In a lens both the surfaces may also be curved. All the distances are measured from the optic centre.The direction of the incident ray is taken as positive and the direction opposite to the incident ray is taken as negative.The refraction formula for spherical curved surface isn2/v -n1/u = n2-n1/r andThe Gaussian formula for a lens is 1/v -1/u = 1/fImages of different sizes and nature are formed by these surfaces of the lens and they depend on the position of the object from the refracting surfaces.The power of a biconvex lens having the raidus of curvature |r1|= |r2| =40 cm of each surface and the refractive index of the material lens n = 3/2 will be(a) 2.5 D(b) 0.25 D(c) 25 D(d) 0.025 D |
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Answer» Correct answer is (a) 2.5 D |
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| 12529. |
There is sudden change is the slopes for some metal oxides like MgO, ZnO , HgO in Ellingham diagram . It is the indication ofA. insufficient supply of oygenB. excess of oxygenC. phase changeD. completion of reaction |
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Answer» Correct Answer - C That is the MP or BP of metal |
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| 12530. |
Explain why `NH_(3)` is basic while `BiH_(3)` is only feebly basic ? |
| Answer» Both N in `NH_3` and `Bi` in `BiH_3` have a lone pair of electrons. Since the atomic size of N is less than Bi and N is more electronegative than Bi Hence electron density is more on N in `NH_3` than on Bi in `BiH_3`. Consequently the tendency to donate its lone pair is more in `NH_3` hence `NH_3` is more basic than `BiH_3`. | |
| 12531. |
100 ml, 0.05 M `CuSO_4` solution is electrolysed by using current of 0.965 Å for 100 min.Find the pH of solution at the end of electrolysis. |
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Answer» Correct Answer - pH=1 pH=1 After 1000 second when total `Cu^2+` discharge then `H^+` and `OH^(-)` discharge equally so pH will not change further. `Cu^(+2)+2e^(-)to Cu` `n_(Cu^(+2))/1=n_(e^(-))/2=Q/(2F)=(it)/(2F)` `(0.925xxt)/(2xx96500)=(0.05xx100)/1000` t=1000 sec `(n_(H^+))_("excess")=(n_(OH^(-)))_("discharge")=n_(e^(-))=Q/F=(it)/(F)=(0.965xx1000)/96500=10^(-2)` `n_(H^+)=(MV)/1000` `10^(-2)=(Mxx100)/1000implies M=10^(-1)` `[H^+]=10^(-1)` `pH= -"log" 10^(-1)=1` |
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| 12532. |
Assertion: Permanganate titrations are not performed in presence of HCIReason: Chlorine is formed as a consequence of oxidation of HCI(a) Both assertion and reason are true, reason is correct explanation of assertion.(b) Both assertion and reason are true, but reason is not a correct explanation of the assertion.(c) Assertion is true, but reason is false(d) Assertion is false, but reason is true |
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Answer» (a) Both assertion and reason are true, reason is correct explanation of assertion. Permanganate titrations are not performed in presence of HCl because HCl oxidises to form chlorine. |
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| 12533. |
Which of the following has the same number of moles as in 398 grams of CuSO4?(a) 2 grams of hydrogen(b) 40 grams of oxygen(c) 35 grams of nitrogen(d) 58.5 grams of Sodium chloride |
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Answer» The correct option is (b) 40 grams of oxygen The explanation is: Number of moles of CuSO4 = 398/ 159 ≈ 2.5, and number of moles of oxygen = 40/16 = 2.5. |
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| 12534. |
What are the number of atoms of chlorine in 250.5 grams of Cl2O6?(a) 1.806 X 10^24(b) 2.408 X 10^24(c) 9.033 X 10^23(d) 6.022 X 10^23 |
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Answer» The correct option is (a) 1.806 X 10^24 Best explanation: Number of moles of Cl2O6 = 250.5/167 = 1.5, => number of atoms of chlorine = 2*1.5*(6.022 X 10^23) = 1.806 X 10^24. |
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| 12535. |
`BrF_(3)` and `IF_(5)` both exist in ionic form, in solid state. What are the hybridizations of cationic parts of both, respectively ?A. `sp^(3), sp^(3)`B. `sp^(3)d, sp^(3)d`C. `sp^(3), sp^(3)d`D. `sp^(2), sp^(3)` |
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Answer» Correct Answer - C `BrF_(3) rarr [BrF_(2)]^(+) [BrF_(4)]^(-)` `IF_(5) rarr [IF_(4)]^(+) [IF_(6)]^(-)` |
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| 12536. |
In solid `N_(2)O_(3)` and solid `N_(2)O_(5),N-O` hybridization in cationic part are respectively:A. `sp^(2)` and `sp`B. `sp` and `sp^(2)`C. `sp` and `sp`D. none of these |
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Answer» Correct Answer - C Solid `N_(2)O_(3) (NO^(-)NO_(3)^(-))` |
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| 12537. |
Which of the following processes involves absorption of energy?A. `S_((g))+e^(_)rarrS_((g))^(-)`B. `O_((g))^(-)+e^(_)rarrO_((g))^(2-)`C. `Cl_((g))+e^(_)rarrCl_((g))^(-)`D. `O_((g))+e^(_)rarrO_((g))^(-)` |
| Answer» Correct Answer - B | |
| 12538. |
In which of the following cases metal obtained by carbon reduction is in liquid state?A. B. C. D. none of these |
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Answer» Correct Answer - C When state of reduced metal changes from solid to liquid and then gas, there is steep increase in value of `DeltaG^(@)`. In case of (1), (2) metal obtained is in gaseous state in case of (3) it is in liquid state. |
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| 12539. |
A neutral atom `(A)` is converted to `(A)` by the following process `Aoverset(E_(1))underset(-e^(-1))rarrA^(+)overset(E_(2))underset(e^(-))rarrA^(+2)overset(E_(3))underset(e^(-))rarrA^(3+)` The correct order of of `E_(1)`, `E_(2)` and `E_(3)` energies is `:-`A. `E_(1)ltE_(2)ltE_(3)`B. `E_(1)gtE_(2)gtE_(3)`C. `E_(1)=E_(2)=E_(3)`D. `E_(1)gtE_(2)ltE_(3)` |
| Answer» Correct Answer - A | |
| 12540. |
When `NaBH_(4)` is dissolved in waterA. `Na^(+)` and `BH_(4)^(-)` ions are formed which are stableB. It decomposes with evolution of `H_(2)`C. `BH_(4)^(-)` is formed initially decomposed to give `H^(ɵ)` ion which prevents further decomposition.D. `NaOH` and `H_(3)BO_(3)` is formed |
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Answer» Correct Answer - B `NaBH_(4)+2H_(2)OrarrNaBO_(2)+4H_(2)uarr` |
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| 12541. |
The radius of isoelectronic series `:-`A. decreases with decreasing nuclear chargeB. decreases with increasing effective nuclear chargeC. same for allD. first increases then decreases |
| Answer» Correct Answer - B | |
| 12542. |
Set of elements having one electron in their valence shell is `:-`A. `Cl,Br,I`B. `Na,Mg,Al`C. `B,Al,Ga`D. `K,Rb,Cs` |
| Answer» Correct Answer - D | |
| 12543. |
In which of the following compounds cation and anion size ratio is minimum?A. `CsF`B. `LiI`C. `LiF`D. `CsI` |
| Answer» Correct Answer - B | |
| 12544. |
For a geseous hydrocarbons, the ratio of volume of `CO_(2)` formed and the volume of `O_(2)` needed for complete combustion is independent from the number of C -atoms. The hydrocarbons isA. AlkaneB. AlkeneC. AlkyneD. Any of these |
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Answer» Correct Answer - B `C_(n)H_(2n) +3nO_(2) to nCO_(2)(g) +nH_(2)O(l)` `(V_(CO_(2))/(V_(O_(2)) =n/(3n) =1/3` which is independent of n |
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| 12545. |
The first ionisation potential of `Na` is `5.1eV`. The value of eectrons gain enthalpy of `Na^(+)` will beA. `-2.55 eV`B. `-5.1` eVC. `-10.2 eV`D. 2.55 eV |
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Answer» Correct Answer - 2 `Na to Na^(+)+e^(-), I.P. =+5.1 eV` `Na^(+) + e to Na to DeltaH_(eg) =-5.1 eV` |
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| 12546. |
(a) growth of body(b) changes in hair pattern(c) change in voice(d) menstruation |
| Answer» (d) menstruation | |
| 12547. |
`H_(2)O_(2)+2KIoverset(40% "yield")rarrI_(2)+2KOH` `H_(2)O_(2)+2KMnO_(4)+3H_(2)SO_(4)overset(50% "yield")rarr K_(2)SO_(4)+2MnSO_(4)+3O_(2)+4H_(2)O` 150mL of `H_(2)O_(2)` sample was divided into two parts. First part was treated with KI and formed KOH required 200 mL of `M//2H_(2)SO_(4)` for neutralisation. Other part was treated with `KMnO_(4)` yielding 6.74 litre of `O_(2)` at 1 atm. and 273 K. Using % yield indicated find volume strength of `H_(2)O_(2)` sample used.A. `5.04`B. `33.6`C. `3.36`D. `11.2` |
| Answer» Correct Answer - D | |
| 12548. |
As per Criminal Procedure Code, 1973 Section 2(x), a warrant case is one that is related to an offence whose punishment can be a death sentence, life imprisonment or imprisonment for a term exceeding ________ years.1. 102. 23. 34. 7 |
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Answer» Correct Answer - Option 2 : 2 The correct answer is 2.
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| 12549. |
Under the Criminal Procedure Code, 1973, the ________ can pass any order to prevent any miscarriage of Justice or abuse of the process in the Indian courts.1. District Court2. Civil Court3. High Court4. Supreme Court |
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Answer» Correct Answer - Option 3 : High Court The correct answer is High Court.
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| 12550. |
The per capita water requirement for hospitals per day is-1. 100-150 Liters2. 150-200 Liters3. 200-250 Liters4. 250-300 Liters |
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Answer» Correct Answer - Option 1 : 100-150 Liters Concept:-
So, per capita, water requirement for hospitals per day is 100-150 Liters. |
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