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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.
| 11851. |
Explain the following terms with suitable examples `:` `a. alc osol`,`b. Aeo rsol`,`c. Hydrosol` |
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Answer» `a.` Alcohol `:` It is a colloidal dispersion having alcohol as the dispersion medium, `e.g.,` collodion `(` a colloidal sol of cellulose nitrate in ethyl alcohol `)`. ltbr. `b.` Aerosol `:` It is a colloidal dispersion of a liquid in a gas, `e.g., `fog. `c.` Hydrosol `:` It is a colloidal sol of a solid in water as the dispersion medium `e.g.,` starch sol or gold sol. |
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| 11852. |
If \( \sin (A+B)=\frac{1}{2} \) and \( \cos (A+B)=\frac{1}{2}, 0^{\circ}B \) then values of \( A \) and \( B \) are |
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Answer» \(sin(A - B) = \frac 12 = sin 30°\) ⇒ A - B = 30° ....(i) \(cos(A + B) = \frac 12 = 60°\) ⇒ \(A + B = 60° \) .....(ii) On adding equations (i) & (ii), we get \((A - B) + (A + B) = 30° + 60° = 90°\) ⇒ \(2A = 90°\) ⇒ \(A = \frac{90°}2 = 45° \) \(\therefore A + B = 60°\) ⇒ \(B = 60° - A = 60° - 45° = 15°\) |
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| 11853. |
\( A=\left[\begin{array}{lll}1 & 0 & 1 \\ 0 & 0 & 0 \\ 1 & 0 & 1\end{array}\right] \) then \( A^{2} \) is equal toAb \( -A \)C \( 2 A \)d \( -2 A \) |
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Answer» Correct option is (c) 2A \(A = \begin{bmatrix}1&0&1\\0&0&0\\1&0&1\end{bmatrix}\) \(A^2 = \begin{bmatrix}1&0&1\\0&0&0\\1&0&1\end{bmatrix} \begin{bmatrix}1&0&1\\0&0&0\\1&0&1\end{bmatrix} \) \( = \begin{bmatrix}2&0&2\\0&0&0\\2&0&2\end{bmatrix}\) \(=2 \begin{bmatrix}1&0&1\\0&0&0\\1&0&1\end{bmatrix}\) \(= 2A\) |
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| 11854. |
\( A=\left[\begin{array}{lll}1 & 0 & 1 \\ 0 & 0 & 0 \\ 1 & 0 & 1\end{array}\right] \) then \( A^{2} \) is equal to |
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Answer» \(A = \begin{bmatrix}1&0&1\\0&0&0\\1&0&1\end{bmatrix}\) \(A^2 = \begin{bmatrix}1&0&1\\0&0&0\\1&0&1\end{bmatrix} \begin{bmatrix}1&0&1\\0&0&0\\1&0&1\end{bmatrix} \) \( = \begin{bmatrix}2&0&2\\0&0&0\\2&0&2\end{bmatrix}\) \(=2 \begin{bmatrix}1&0&1\\0&0&0\\1&0&1\end{bmatrix}\) \(= 2A\) |
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| 11855. |
ix) \( \sin ^{3} x \) |
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Answer» \(sin^3 (2\pi + x) = (sin(2\pi + x))^3 = (sin \, x)^3 = sin^3 x\) \(\therefore \) Period of sin3x is 2π. |
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| 11856. |
Closer property |
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Answer» closer property Addition : The sum of any two integers is also an integer. Subtraction : The difference of any two integers is also an integer. Multiplication : The product of any two integers is also an integer. Division : The division of any two integers not always an integer.
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| 11857. |
यदि `AB = (1)/(2): (3)/(8), B : C = (1)/(3) : (5)/(9)` और `C : D = (5)/(6) : (3)/(4)` है, तो `A : B : C: D` का अनुपात ज्ञात करे ?A. `6 : 4 : 8 : 10`B. `6 : 8 : 9 : 10`C. `8 : 6 : 10 : 9`D. `4 : 6: 8 : 10` |
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Answer» Correct Answer - C `A : B = (1)/(2) : (3)/(8), (A)/(B)=(4)/(3)` `B : C = (1)/(3) : (5)/(9)` `(B)/(C) = (1)/(3) xx(9)/(5) = (3)/(5)` `C : D = (5)/(6):(3)/(4),(C)/(D) = (5xx4)/(6xx3)=(10)/(9)` `A : B : C : D` `4 : 3` `3 : 5` `10 : 9` `rArr {:(A,:,B,:,C,:,D),(8,:,6,:,10,:,9):}` |
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| 11858. |
Determine whether the points ଵ (−6, 4, 8), ଶ(9, −2, 0) and ଷ(1, −5, 3) lie on the same line. |
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Answer» A = (-6,4,8), B = (9,-2,0), C = (1, -5,3) AB = \(\sqrt{(9-(-6))^2+(-2-4)^2+(0-8)^2}\) = \(\sqrt{15^2+(-6)^2+(-8)^2}\) = \(\sqrt{225+36+64}\) = \(\sqrt{325}\) = \(5\sqrt13\) = 18.028 BC = \(\sqrt{(1-9)^2+(-5+2)^2+(3-0)^2}\) = \(\sqrt{8^2+3^2+3^2}\) = \(\sqrt{64+9+9}\) = \(\sqrt{82}\) = 9.055 AC = \(\sqrt{1-(-6))^2+(-5-4)^2+(3-8)^2}\) = \(\sqrt{49+81+25}\) = \(\sqrt{155}\) = 12.449 Clearly, AB \(\ne\)BC + AC Or AC \(\ne\) AB + BC Or BC \(\ne\) AB + AC Therefore, points A, B and C are not collinear(not on same plane) ,they forms a triangle. |
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| 11859. |
The value of k such that the line (x - 4)/1 = (y - 2)/1 = (z - k)/2 lies in the plane 2x - 4y + z - 7 = 0 is(A) 7 (B) −7 (C) no real value (D) 4 |
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Answer» Correct option is (A) 7 Explanation : Every point on the given line is of the form (4 + λ, 2 + λ, k + 2λ), where λ ∈ R. This point lies in the plane 2x - 4y + z - 7 = 0 So 2(4 + λ) - 4(2 + λ) + l(k + 2λ) - 7 = 0 for all λ ∈ R In particular, for λ = 0, we have 8 - 8 + k - 7 = 0 k = 7 Note that the point (4, 2, k) lies on the given line and hence it lies in the given plane. So 2(4) - 4(2) + k - 7 = 0 k = 7 |
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| 11860. |
The plane 2x + y + 2z − 9 = 0 meets the coordinate axes at points A, B and C. Then the orthocentre of triangle ABC is(A) 3/2,3,3/2 (B) (2, 1, 2) (C) 9/7,27/7,9/7(D) 3/13,1/13,3/13 |
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Answer» Correct option is (B) (2, 1, 2) |
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| 11861. |
A mirror and a source of light are situated at origin O & at a point on OX respectively. A ray of light from sources strikes the mirror along x-axis & reflected. If the DRs of normal to plane are `lt1-1,1gt`, then DCs of reflected ray areA. `(:(1)/(3),(2)/(3),(2)/(3):)`B. `(:(-1)/(3),(-2)/(3),(2)/(3):)`C. `(:(-1)/(3),(-2)/(3),(-2)/(3):)`D. None |
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Answer» Correct Answer - B For B to win: PR or `overset(RBR)(BR R)` `{:("Black",,"Red",,"B(n)",,,),(1,,2,,B(1)=1,,,overset(A(n))A(1)=0),(2,,2,,B(2)=(2)/(4).(1)/(3)+(2)/(4).(2)/(3).(1)/(2)+(2)/(4).(2)/(3).(1)/(2),,,),(3,,2,,B(3)=(2)/(5).(1)/(4)+(2)/(5).(3)/(4).(1)/(3)+(3)/(5).(2)/(4).(1)/(3),,,),(n,,2,,B(n)=(2)/(n+2).(1)/(n+1)+(2)/(n+2).(n)/(n+1).(1)/(n)+(n)/(n+2).(2)/(n+2).(1)/(n),,,):}` `therefore B(n) = (6)/((n+1)(n+2))` `A(n)=1-B(n)=((n-1)(n+4))/((n+1)(n+2))` `pi[A(n)]=(6xx(n+2)(n+3)(n+4))/(3xx4xx5.n(n+1)(n+2))` |
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| 11862. |
A plane intersects the coordinate axes in A, B and C, respectively. If (α,β,γ) is the centroid of Δ ABC, then show that the equation of the plane x/α + y/β + z/γ = 3 |
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Answer» Let A = (a,0,0), B = (0,b,0) and C = (0,0,c) so that G = (a/3, b/3, c/3) and the equation of the plane is x/a + y/b + z/c = 1 ...(1) Since (α, β, γ) is the centroid, we have α = a/3, β = b/3, γ = c/3 a = 3α, b = 3β, c = 3γ so that from Eq. (1), the equation of the planes is x/α + y/β + z/γ = 3 |
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| 11863. |
The planes 2x + y + z + 4 = 0, 3x + 2y + z + 8 = 0 and y − z + 4 = 0(A) do not have common points. (B) have a unique common point. (C) form a triangular prism. (D) intersect in a line |
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Answer» Correct option is (D) intersect in a line. |
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| 11864. |
The vertices of a triangle ABC are A(−1, −2, −3), B(−1, 2, 3) and O(0, 0, 0). The direction ratios of the internal bisector of the angle AOB are(A) 0, 0, 1 (B) −1, 0, 0 (C) −1, 1, 1 (D) −1, −1, −1 |
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Answer» Correct option is (B) −1, 0, 0 |
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| 11865. |
कर्नाटक का दूसरा युद्ध किनके बीच हुआ था ? |
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Answer» कर्नाटक का दूसरा युद्ध मुजफ्फरजंग तथा नासिरजंग के बीच हुआ था। |
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| 11866. |
आधुनिक साहित्य की सबसे अधिक लोकप्रिय विधा का नाम लिखिए। |
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Answer» आधुनिक साहित्य की सबसे अधिक लोकप्रिय विधा का नाम ‘कहानी’ है। |
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| 11867. |
Ammonium ion is- (A) A Lewis acid (B) Lewis base (C) Bronsted acid (D) Bronsted base |
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Answer» Ammonium ion is Lewis base. |
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| 11868. |
For a body rotating with constant angular velcoity, its kinetic energy is directly proportional to the square of its |
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Answer» Angular momentum of a body , `L = Iomega` ( in magnitude) `therefore` Rotational kinetic energy of a body. `E = (1)/(2)Iomega= (1)/(2)(Iomega)omega = (1)/(2)Lomega = (1)/(2)L((L)/(1))= (1)/(2)(L^(2))/(I)` |
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| 11869. |
An ideal mono atomic gas is taken through the cyclic process shown in fig. Linear expansion from A to B following by adiabatic compression back to original state A. Efficiency of cyclic process is zeroB. During process A to B heat is released onlyC. During process A to B heat is absorbed onlyD. During linear process heat enters and leaves the process |
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Answer» Correct Answer - D Since `P_(1)V_(1) gt P_(2)V_(V) rArr T_(A) gt T_(B)` work done is `+ve` and efficiency is positive quantity `eta = ("work done")/("Heat given") = 1 - (Q_("released"))/(Q_("absorbed"))` Therefore there must be some heat absorbed in the process. During adiabatic process `DeltaQ = 0`, therefore during linear process heat will entre and leaves system at different times. |
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| 11870. |
What two changes on the equilibrium N2(g) + 3H2(g) ⇄2NH3(g), ΔH = –92.5 kJ can keep its state undisturbed ? |
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Answer» Increase of temperature alongwith suitable increase of pressure or increase of presure alongwith suitable increase of temperature. |
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| 11871. |
if density of an ideal gas when plotted against pressure exerted, shows the above variation at 273K. Find the molar mass (in gm/mol) the gas?A. 19.4B. 38.8C. 77.6D. 100 |
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Answer» PM=dRT `d=P((M)/(RT))` slope `=tan 60^(@)=(M)/(RT)` `M=sqrt(3)xx0.0821xx273approx39gm` |
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| 11872. |
The specific heat of a polyatomic gas is greater than the specific heat of a monatomic gas because: A. the polyatomic gas does more positive work when energy is absorbed as heat B. the monatomic gas does more positive work when energy is absorbed as heat C. the energy absorbed by the polyatomic gas is split among more degrees of freedom D. the pressure is greater in the polyatomic gas E. a monatomic gas cannot hold as much heat |
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Answer» C. the energy absorbed by the polyatomic gas is split among more degrees of freedom |
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| 11873. |
Monatomic, diatomic, and polyatomic ideal gases each undergo slow adiabatic expansions from the same initial volume and the same initial pressure to the same final volume. The magnitude of the work done by the environment on the gas: A. is greatest for the polyatomic gas B. is greatest for the diatomic gas C. is greatest for the monatomic gas D. is the same only for the diatomic and polyatomic gases E. is the same for all three gases |
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Answer» A. is greatest for the polyatomic gas |
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| 11874. |
Which of the following change at equilibrium will shift reaction in backward direction: `Fe^(3+)(aq)+SCN^(-)(aq)hArrFe(SCN)^(2+)(aq)`A. Addition of waterB. Addition of `KOH(aq)[Fe(OH)_(3)` is insoluble in water]C. Addition of `NaNO_(3)(s)`D. A and B both |
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Answer» `Fe^(3+)(aq)+SCN^(-)(aq)hArrFe(SCN)^(2+)(aq)` (A) For given Q `alpha` V `rArr` on addition of water `QgtK` reaction move backward (B) `Fe^(+3)+3OH^(-)hArrFe(OH)_(3)(s)` reaction moves backward |
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| 11875. |
An electric heater supplies heat to a system at a rate of 100 W. If system performs work at a rate of 75 Joules per second. At what rate is the internal energy increasing? |
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Answer» Given ∆Q = 100w = 100J/s ∆W = 75 J/s ∆Q = ∆U + ∆W ∆U = ∆Q – ∆W = 100 – 75 = 25J/S |
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| 11876. |
A heat engine is a device which converts heat energy into work.1. What is the working substance in an ideal heat engine.2. A Carnot engine is working between in melting point and steam point. What is its efficiency? |
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Answer» 1. Ideal gas. 2. T2 = 0°C = 273K T1 = 100°C = 373K Efficiency η = I – \(\frac{T_2}{T_1}\) = I−\(\frac{273}{373}\) η = 0.27 η = 27%. |
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| 11877. |
What amount of heat must be supplied to 2.0 x 10-2 kg of nitrogen (at room temperature) to raise its temperature by 45°C at constant pressure? (Molecular mass of N2 = 28, R = 8.3 J mol-1 K-1). |
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Answer» m = 2 × 103 kg-1 = 20g, ∆T = 45°C, M = 28 R = 8.3 J mol-1 K-1, M = 28 Number of moles, n = \(\frac{m}{M} = \frac{20}{5} = \frac{5}{7}\) Since nitrogen is diatomic, ∴ CP = \(\frac{7}{2}\)R = \(\frac{7}{2}\) × 8.3 J mol-1K-1 Now, ∆Q = nCP ∆T \(\frac{5}{2} \times \frac{7}{2}\)× 8.3 × 45J = 933.75J. |
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| 11878. |
Identify the thermodynamic process in which temperature of system may increase even when no heat is supplied to the system. |
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Answer» In Adiabatic process temperature of system may increase even when no heat is supplied to the system. |
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| 11879. |
The door of an operating refrigerator kept open in a closed room. Will it make the room cool? |
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Answer» No. The room will be slightly warmed. |
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| 11880. |
Which thermodynamic variable is defined by first law of thermodynamics? |
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Answer» Internal energy variable is defined by first law of thermodynamics. |
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| 11881. |
Which of the following laws of thermodynamics forms the basis for the definition of temperature? (a) First law (b) Zeroth law (c) Second law (d) Third’law |
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Answer» (b) Zeroth law The Zeroth law of thermodynamics gives the definition of temperature. |
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| 11882. |
In air-conditioning design for summer months, the condition inside a factory where heavy work is performed as compared to a factory in which light work is performed should have (a) lower dry bulb temperature and lower relative humidity (b) lower dry bulb temperature and higher relative humidity (c) lower dry bulb temperature and same relative humidity (d) same dry bulb temperature and same relative humidity. |
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Answer» (a) lower dry bulb temperature and lower relative humidity |
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| 11883. |
Zeroth Law of thermodynamics states that (a) two thermodynamic system are always in thermal equilibrium with each other. (b) if two systems are in thermal equilibrium, then the third system will also be in thermal equilibrium. (c) two systems not in thermal equilibrium with a third system are also not in thermal equilibrium with each other. (d) when two systems are in thermal equilibrium with a third system, they are in thermal equilibrium with each other. |
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Answer» (d) when two systems are in thermal equilibrium with a third system, they are in thermal equilibrium with each other. |
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| 11884. |
A gaseous mixture of composition by volume, 78% H2 and 22% CO is contained in a vessel. It is desired that the mixture should be made in proportion 52% H2 and 48% CO by removing some of the mixture and adding some CO. Calculate per mole of mixture the mass of mixture to be removed, and mass of CO to be added. Assume that the pressure and temperature in the vessel remain constant during the procedure. |
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Answer» Composition of mixture by volume : H2 = 78%, CO = 22% Final composition desired : H2 = 52%, CO = 48% Since the pressure and temperature remain constant, then the number of moles in the vessel remain the same throughout. ∴ Moles of mixture removed = Moles of CO added. Let x kg of mixture be removed and y kg of CO be added. For the mixture, M = \(\sum\cfrac{V_i}VM_i\) M = 0.78 × 2 + 0.22 × 28 = 7.72 Also from equation, n = \(\cfrac mM\), we have Moles of mixture removed = \(\cfrac{x}{7.72}\) = moles of CO added = \(\cfrac y{28}\) From equation, \(\cfrac{V_i}V\) = \(\cfrac{n_i}n\), we have Moles of H2 in the mixture removed = 0.78 × \(\cfrac{x}{7.72}\) = 0.101 x Moles of H2 initially = 0.78 × 1 = 0.78 Hence, Moles of H2 remaining in vessel = 0.78 – 0.101 x But 1 mole of the new mixture is 52% H2 and 48% CO, therefore 0.78 – 0.101 x = 0.52 ∴ 0.101 x = 0.26 or x = 2.57 Mass of mixture removed = 2.57 kg Also since \(\cfrac{x}{7.72}\) = \(\cfrac y{28}\) y = \(\cfrac{28}{7.72}\) \(\times\) x = \(\cfrac{28}{7.72}\) x 2.57 = 9.32 Mass of CO added = 9.32 kg. |
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| 11885. |
A particle free to move along the (x - axis) hsd potential energy given by `U(x)= k[1 - exp(-x^2)] for -o o le x le + o o`, where (k) is a positive constant of appropriate dimensions. Then.A. at points away from the origin, the particle is in unstable equilibrium.B. for any finite non-zero value of `x`, there is a force directed away from the originC. if its total mehanical energy is `k//2`, it has its minimum kinetic energy at the originD. for small displacements from `x = 0`, the motion is simple harmonic. |
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Answer» Correct Answer - D Let us plot the graph of the mathematical equation `U(x) = K[1- e^(-x^(2))]` `:. F = -(dU)/(dx) =-2kxe^(-x^(2))` It is clear that the potential energy is minimum at `x =0`. Therefore, `x = 0` is the state of stable equilibrium. Now if we displace the particle from `x = 0`, then for small displacements the particle tends to regain the position `x = 0` with a force `F = 2kx//e^(x^(2))` for `x` to be small `F prop x`. |
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| 11886. |
Who circulated a questionnaire on Uniform Civil Code in October, 2016 to solicit options from the public at large?1. Ministry of Law and Justice2. National Commission for Minorities3. National Human Rights Commission4. National Law Commission |
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Answer» Correct Answer - Option 4 : National Law Commission The correct answer is the National Law Commission.
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| 11887. |
The work done by a force on a body moving from (1,4) m to (3,8) m is 44 J. The force can be :A. `8 hat i+6hatj N`B. `8 hati-8 hatjN`C. `6 hati-8 hatj N`D. `6 hati+8 hatj N` |
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Answer» Correct Answer - D `W=vec(F).vec(S)` `44=vec(F)(2hati+4hatj)` by hit and trial `(6hati+8hatj).(2hati+4hatj)=44 J` |
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| 11888. |
The upwards normal force exerted by the force of an elevator on a passenger who weighs 600 N is 570 N. The acceleration of passanger isA. ZeroB. `1/2 m//s^(2) uarr`C. `1/2 m//s^(2) darr`D. `10 m//s^(2) darr` |
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Answer» Correct Answer - C Since, NgtW `:.` acceleration (a) is downwards Also N=m(g-a) `:. 570=60(g-a)` `rArr a=1/2 m//s^(2) (darr)` |
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| 11889. |
The Bhakra Nangal project is on the river1. Damodar2. Sutlej3. Mahanadi4. Tungabhadra |
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Answer» Correct Answer - Option 2 : Sutlej The correct answer is Sutlej.
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| 11890. |
Narmada and Tapi flow into the (a) Gulf of Khambat (b) Palk Strait(c) Bay of Bengal (d) Gulf of Kachchh |
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Answer» Narmada and Tapi flow into the Gulf of Khambat. |
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| 11891. |
Narmada and Tapi flow into the (a) Gulf of Khambat (b) Palk Strait (c) Bay of Bengal (d) Gulf of Kachchh |
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Answer» Narmada and Tapi flow into the Gulf of Khambat. |
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| 11892. |
Two pistons having low thermal conductivity divide an adiabatic container in three equal parts as shown An ideal gas is present in the three parts AB&C having initial pressures as shown and same temperatures Now the pistons are releasef Then the final equilibrium length of part A after long time will be A. `L//8`B. `L//4`C. `L//6`D. `L//5` |
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Answer» Correct Answer - B `n=(PV)/(RT)` Final pressure & temperature is same hence volume will be in ratio of number of moles. |
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| 11893. |
23. Force position relation of particle executing SHM on \( x \)-axis is \( F+2 x-1=0 \) dyne, where \( x \) in centimeter. Mass of particle is \( 8 gm \), then find out :(i) Mean position(ii) Time period of one oscillation |
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Answer» F = - 2x + 1 F = -2 \((x - \frac{1}{2})\) F = \(-2 \times 10^{-5} \times 10^{-2}(x - \frac{1}{2})\) F = \(-2 \times 10^{-7}(x - \frac{1}{2})\) F = - kx k = \(2 \times 10^{-7}\) We know that \(\omega = \sqrt{\frac{k}{m}}\) \(\omega = \sqrt{\frac{2 \times 10^{-7} \times 1000}{0.008}}\) \(\omega = \sqrt{0.25 \times 10^{-4}}\) \(\omega = 0.5 \times 10^{-2}\) rad/sec (i) Mean position zero. (ii) Time period \(T = \frac{2\pi}{\omega}\) \(=\frac{2\pi}{0.5 \times 10^{-2}}\) \(T = \frac{20\pi \times 10^2}{5}\) \(T = 4\pi \times 10^2 \, sec\) |
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| 11894. |
The displacement equation of SHM is y = 0.5 (sin 3πt + √3 cot 3πt). Find Amplitude, Angular frequency, Period of oscillation, velocity and acceleration. |
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Answer» y = 0.5 (sin 3πt + √3 cot 3πt) y = 0.5 x 2 ( \(\cfrac12\) sin 3πt + \(\cfrac{\sqrt3}2\) cot 3πt) y = 1 ( cot \(\cfrac{\pi}3\) sin 3πt - sin \(\cfrac{\pi}3\) cot 3πt) y = sin( 3πt + \(\cfrac{\pi}3\)) Comparing equation y = A sin (ωt + θ) Then amplitude A = 1 Angular frequency ω = 3π rad/sec Time period T = \(\cfrac{2\pi}ω\) T = \(\cfrac{2\pi}{3\pi}\) T = \(\cfrac23\) sec Acceleration a = Aω2 = 1 x (3π)2 a = 9π2 m/s2 Velocity v = Aω v = 1 x 3π v = 3π m/s |
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| 11895. |
A cube and a sphere (both solid) of same mass, same material are heated upto same temperature and then kept in same surrounding. Rate of cooling means rate of fall of temperatureA. Sphere will cool down at a faster rate.B. cube will cool down at a faster rate.C. ratio of rate of cooling of cube to sphere `=2[3/(4pi)]^(1//3)`D. ratio of rate of cooling of cube to sphere =`1/2 (3/(4pi))^(1//3)` |
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Answer» Correct Answer - BC `(dT)/(dt)=(CA)/(ms)(T-T_(0))` `l^(3) P=4/3 pi R^(3) P` `R=sqrt(3/(4pi))l` `(dT_(C)//dt)/(dT_(s)//dt)=(A_(C))/(A_(S))=(6l^(2))/(4piR^(2))` `=(6l^(2))/(4pixx(3//4pi)^(2//3) l^(2))=2sqrt(3/(4pi))` |
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| 11896. |
A sine wave y=sin `(2pix-2pit+pi//3)` is propagating in the medium. Match the description of the motion of particles of the medium with entries in columnI `{:(,"Column-I",,"Column-II"),("(A)",x=1/3"m, t"=1/3sec,,"(P)Velocity is in positive y direction"),("(B)",x=1/3"m, t"=1sec,,"(Q)Velocity is in negative y direction"),("(C)",x=1"m, t"=1/3sec,,"(R)Particle is stationary"),("(D)",x=1"m, t"=1sec,,"(S)Particle has positive displacement"),(,,,"(T)Particle has negative displacement"):}` |
| Answer» Refrection at movable boundaries no phase reversal takes palce but at rigid bondaries phase reversal of `180^(@)` takes place | |
| 11897. |
A wave y= 3mm `sin (2pix-200 pit)` is Propagating in the left string. It strikes the boundary at x=0 and is reflected partially and tranmitted partially. Choose the corrects statements (s)A. The tension in strings is 10NB. The amplitude of transmitted wave is 2 mm.C. The wave undergoes a phse change of `pi` on reflection.D. The wavelength of transmitted wave is `1/2` m |
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Answer» Correct Answer - ABCD `(A) v=(omega)/k =100 m//s =sqrt(T/(10^(-3)))` `rArr T=10 N` `(B) A_(t) =(2V_(2))/(V_(1)+V_(2)) Ai=(2xx50)/150 xx3=2 mm` (D) `lambdaf=v`, v is halfed so `lambda` is also halfed. |
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| 11898. |
The equation of the progressive wave is y =a `sin pi (nt -(x)/(5))` the ratio maximum paritcle velocity to wave velocity isA. `(pi a)/(5)`B. `(2pi a)/(5)`C. `(3pi a)/(5)`D. `(4pia)/(5)` |
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Answer» Correct Answer - c We have y = a sin `2pi(nt-(x)/(5))` here vi is wave velocity `u_(max)/(v)=(2pia)/(lambda)=(2pia)/(5)[therefore k=(2pi)/(lambda)]` `lambda=(2pi)/(k)=(2pi)/(2pi//5)=5` |
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| 11899. |
Initially spring is in natural length and block of mass m is released from rest what should be the minimum value of M (in kg) to just move the 2kg block. If this minimum value is `(x)/(8)`kg. Find x |
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Answer» Correct Answer - 6 With respect to free end, velocity of block is 3 m/s leftward. In position of maximum extension, the block will stop w.r.t. free end. As observed from free end, which is inertial frame `(1)/(2)kx^(2)=(1)/(2)mv^(2)` `x=sqrt((mv^(2))/(k))=3sqrt((m)/(k))=3sqrt((1)/(10^(4)))=3xx10^(-2)` meter = 3 cm |
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| 11900. |
Hot soup tastes better than cold soup. Why? |
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Answer» The surface tension of hot soap is less compared to cold soap. So hot soap spreads larger area. |
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