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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. |
Atead of teknik te madh eseerveral dit ad miongue wharate dewite Mestrel is soomla with working stres ADSL11 des take osno0nn from2) TOP3) TOPPr 100 x 10ros = 5oТе са= 4dla PL - 1000 10x29103AE (1000 qq) x 200x10²1.zumA short hollow I cylinder of wall thickness of 10 mon & tocarry a compressive load of 600kN is outside dia if theultimate stress is suompa take factor of sabely 24? 2600x10"max load 2540546 - 135Dad+24Dod+206067d 2131.4me |
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Answer» Answer: Introduction: Books are our best companies because it gives us knowledge and peace of mind unconditionally. Although we have friends and family to share our thoughts and get suggestions, we think books are the best when it comes to live a life full of wisdom, but no explanation. A book can be the one thing which I can read all day long leaving behind all the problems. I like books because they give us knowledge, but never expect us, or force us to remain obedient to one proven or unproven piece of information. Many REASONS to Love Books: These are not the only reasons that I think books are our best companies. Books also offer knowledge about different aspects of life. If you’ve to understand a concept in depth, read books. Books are well KNOWN for accurate and impartial information. Also, books are mostly written by a subject expert. This assures detailed research and information about a subject when you read a book. This is why books are always good when you’ve to get a real understanding about something. Books are the most easily accessible source of knowledge as well. If you want to read books, you can go to a LIBRARY and get a book you want. I know there are faster modes of accessing information like the internet, but the internet is mainly for finding information, not knowledge. Furthermore, only a very few websites or blogs can be trusted to rely on the information you get. The internet is full of wrong information which can mislead us. This is why books are always the best option. Books are Our Best Companies: Another reason I think books are our best companies because books can keep us productively engaged. What I mean to say is READING a book is a very good habit and also a time pass. When you read books, you develop a strong concentration power. To read a book and understand the facts, you’ve to read it uninterruptedly. This develops a habit of reading and an ability to concentrate for very long hours. Books are our best companies because this is an impartial knowledge sharing process. If you’re reading books on a specific subject, you can refer to many books written on the same subjects, but by different authors. This way you can get an idea about the different opinions of the experts, their thoughts processes and beliefs. You can then decide to follow or not to follow an IDEOLOGY. The knowledge and information you get from books are therefore always impartial and more extensive. HOPE it HELPS you.... MARK me as branlist ❤❤ |
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| 2. |
Two identical parallel plate capacitors, of capacitance C each, have plates of area A, separated by a distance d. The space between the plates of the two capacitors, is filled with three dielectrics, of equal thickness and dielectric constants K₁, K₂ and K₃. The first capacitor is filled as shown in fig. I, and the second one is filled as shown in fig. II. If these two modified capacitors are charged by the same potential V, the ratio of the energy stored in the two, would be (E₁refers to capacitor (I) and E₂to capacitor (II)) :options |
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Answer» E1/E2 = 9k1k2k3/(k1+K2+k3)[k2k3+k3k1+k1k2] Option B is correct •When a dielectric of dielectric CONSTANT K is inserted between two plates of CAPACITOR then : C' = KC ______(1) • I) Arrangement will behave as series combination of the capacitor each with capacitance kAE/(d/3) i.e. 3kAE/d where , k is dielectric constant HENCE net capacitance will be 1/C = 1/C1 + 1/C2 + 1/C3 1/C = 1/(3k1AE/d) + 1/(3k2AE/d) + 1/(3k3AE/d) 1/C = d/3AE [ 1/k1 + 1/K2 + 1/K3] 1/C = d[k2k3+k3k1+k1k2]/3AE.[k1k2k3] C = 3AE[k1k2k3]/d[k2k3+k3k1+k1k2] •E1 = CV²/2 •E1 = 3AE[k1k2k3]V²/2d[k2k3+k3k1+k1k2] _______(2) •Similarly II) Arrangement will behave as parallel combination of the capacitor each with capacitance k(A/3)E/d i.e. kAE/3d where , k is dielectric constant Hence net capacitance will be C = C1 + C2 + C3 C = k1AE/3d + k2AE/3d + k3AE/3d C = AE( k1 + k2 + k3 )/3d E = CV²/2 E2 = AE( k1 + k2 + k3 )V²/3(2)d _______(3) Dividing (2) &(3) •E1/E2 = {3AE[k1k2k3]V²/2d[k2k3+k3k1+k1k2]}/ AE( k1 + k2 + k3 )V²/3(2)d •E1/E2 = 9k1k2k3/(k1+k2+k3)[k2k3+k3k1+k1k2] |
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| 3. |
A particle starts from origin O from rest and moves with a uniform acceleration along the positive x – axis. Identify all figures that correctly represent the motion qualitatively. (a = acceleration, v = velocity, x = displacement, t = time)(A) a, b, c (B) a(C) b, c (D) a, b, d |
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Answer» THANK YOU Explanation: |
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| 4. |
A particle P is formed due to a completely inelastic collision of particles x and y having de – Broglie wavelengths λₓ and λᵧ respectively. If x and y were moving in opposite directions, then the de – Broglie wavelength of P is:(A) λₓ + λᵧ (B) (λₓλᵧ) /(λₓ+ λᵧ)(C) (λₓλᵧ) /|λₓ - λᵧ|(D) λₓ - λᵧ |
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Answer» <P>Answer: Explanation: By momentum conservation P X −P y =P p h / λ x - h / λ y = h / λ p (λₓλᵧ) /|λₓ - λᵧ| HOPE THAT IT WAS HELPFUL!!!! MARK IT THE BRAINLIEST IF IT REALLY WAS!!!
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| 5. |
In Li⁺⁺, electron in first Bohr orbit is excited to a level by a radiation of wavelength λ. When the ion gets deexcited to the ground state in all possible ways(including intermediate emission) a total of six spectral lines are observed. What is the value of λ?(Given: h = 6.63 × 10³⁴ js; e = 3 × 10⁸ ms⁻¹)(A) 10.8 nm (B) 11.4 nm(C) 9.4 nm (D) 12.3 nm |
Answer» answer : OPTION (A) 10.8 nmgiven, total number of SPECTRAL lines are observed = 6 ⇒n(n - 1)/2 = 6 ⇒n² - n - 12 = 0 ⇒n² - 4n + 3n - 12 = 0 ⇒n = 4, - 3 so, n = 4 hence transmission from n₁ = 1 to n₂ = 4 now ENERGY , ∆E = 13.6 × Z²[1/n₁² - 1/n₂² ] = 13.6 × (3)² [1/1² - 1/4²] = 13.6 × 9 × 15/16 = 114.75 eV now using FORMULA, ∆E = hc/λ λ = 1240/∆E (in eV) nm = 1240/114.75 = 10.8 nm hence option (A) is correct choice.. |
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| 6. |
A uniform cable of mass ‘M’ and length ‘L’ is placed on a horizontal surface such that its (1/n)th part is hanging below the edge of the surface. To lift the hanging part of the cable upto the surface, the work done should be :(A) nMgL (B) MgL/2n²(C) 2MgL/n²(D) MgL/n² |
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Answer»
hi mate....♥ A UNIFORM cable of mass M and LENGTH L is placed on a horizontal surface such that its 1/n TH parth is HANGING below the edge of the surface. To lift the hanging part of the cable upto the surface, the work done should be: a) MgL/n² b) MgL/2n² c) 2MgL/n² d) nMgL hope it was helpful ♥ |
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| 7. |
A man (mass = 50 kg) and his son (mass = 20 kg) are standing on a frictionless surface facing each other. The man pushes his son so that he starts moving at a speed of 0.70 ms⁻¹ with respect to the man. The speed of the man with respect to the surface is :(A) 0.47 ms⁻¹(B) 0.28 ms⁻¹(C) 0.14 ms⁻¹(D) 0.20 ms⁻¹ |
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Answer» (B) the answer is B. follow me |
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| 8. |
The temperature, at which the root mean square velocity of hydrogen molecules equals their escape velocity from the earth is closest to:[Boltzmans Constant ᵏB = 1.38 10⁻²³ J/K Avogadro number Nₐ = 6.02 × 10²⁶ /kgRadius of Earth: 6.4 × 10⁶ mGravitation acceleration on Earth = 10 ms⁻²] (A) 800k (B) 10⁴ K(C) 3 ×10⁵ (D) 650 K |
Answer» B) 10⁴ KThe rms velocity of Hydrogen will be equal to their escape velocity from Earth at a TEMPERATURE of 10⁴ K
Boltzmann Constant K Avogadro number Nₐ = 6.02 × 10²⁶ /kg Radius of Earth: R Gravitation acceleration on Earth = G = 10 ms⁻²
M = 2 kg ........................(2)
Substituting the values in (5), we get T ≈ 10 × 10³ = 10⁴ Kelvin |
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| 9. |
Diameter of the objective lens of a telescope is 250 cm. For light of wavelength 600 nm. Coming from a distant object, the limit of resolution of the telescope is close to:-(A) 1.5 × 10⁻⁷ rad (B) 2.0 × 10⁻⁷ rad (C) 3.0 × 10⁻⁷ rad (D) 4.5 × 10⁻⁷ rad |
Answer» Answer:Correct OPTION is : ( C ). 3.0 × 10⁻⁷ radEXPLANATION :Refer the ATTACHED picture. |
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| 10. |
A spaceship orbits around a planet at a height of 20 km from its surface. Assuming that only gravitational field of the plant acts on the spaceship. What will be the number of complete revolutions made by the spaceship in 24 hours around the plane?[Given: Mass of plane = 8 × 10²² kg, Radius of planet = 2 × 10⁶ m, Gravitational constant G = 6.67 × 10⁻¹¹ Mn²/kg²](A) 9 (B) 11(C) 13 (D) 17 |
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Answer» Explanation: Given A spaceship orbits around a PLANET at a height of 20 km from its surface. Assuming that only gravitational field of the plant acts on the spaceship. What will be the number of complete REVOLUTIONS made by the spaceship in 24 hours around the plane?
Reference link will be |
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| 11. |
The formula X = 5YZ² X and Z have dimensions of capacitance and magnetic field respectively. What are the dimensions of Y in SI units?(A) [M⁻² L⁰ T⁻⁴ A⁻²] (B) [M⁻³ L⁻² T⁸ A⁻¹](C) [M⁻² L⁻² T⁶ A³] (D) [M⁻¹ L⁻² T⁴ A²] |
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Answer» The dimensions of Y in SI UNITS is : M⁻³.L⁻².T⁸.A⁴ The dimension can be FOUND out in the FOLLOWING ways- Given, X = 5YZ² , therefore,
Thus, the dimension of Y in SI unit is M⁻³.L⁻².T⁸.A⁴. |
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| 12. |
Young’s moduli of two wires A and B are in the ratio 7 : 4. Wire A is 2 m long and has radius R. Wire A is 2 m long and has radius R. Wire B is 1.5 m long and has radius 2 mm. If the two wires stretch by the same length for a given load, then the value of R is close to:(A) 1.3 mm (B) 1.5 mm(C) 1.7 mm (D) 1.9 mm |
Answer» givenlength of wire A = 2 m radius of wire A = R and length of wire B = 1.5 m radius = 2mm YOUNG modulus of A : young modulus of B = 7:4 STEP by step explanation see the attachment
correct OPTION is c) 1.7 MM |
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| 13. |
Following figure shows two processes A and B for a gas. If ∆QA and ∆QB are the amount of heat absorbed by the system in two cases, and ∆UA and ∆UB are changes in internal energies, respectively, then:(A) ∆QA = ∆QB; ∆UA = ∆UB(B) ∆QA > ∆QB; ∆UA = ∆UB(C) ∆QA < ∆QB; ∆UA < ∆UB(D) ∆QA > ∆QB; ∆UA > ∆UB |
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Answer» Answer: (B). will be correct ANSWERS HEAT is a path dependant process .and in process 1st volume change is more than process 2nd. Internal is state FUNCTION and in both process initial and final state is same .so change in internal energy will be zero. Explanation: hope the answer will help you Mark the answer brainliests please. |
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| 14. |
In a simple pendulum experiment for determination of acceleration due to gravity (g), time taken for 20 oscillations is measured by using a watch of 1 second least count. The mean value of time taken comes out to be 30s. The length of pendulum is measured by using a meter scale of least count 1 mm and the value obtained is 55.0 cm. The percentage error in the determination of g is close to:(A) 0.7 % (B) 3.5%(C) 6.8% (D) 0.2% |
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| 15. |
A convex lens of focal length 20 cm produces images of the same magnification 2 when an object is kept at two distances x₁ and x₂ (x₁ > x₂) from the lens. The ratio of x₁ and x₂ is:(A) 5 : 3 (B) 2 : 1(C) 4 : 3 (D) 3 : 1 |
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Answer» Answer: HELLO! Explanation: Magnification is 2 If image is REAL, x 1
= 2 3f
If image is virtual, x 2
= 2 f
x 2
x 1
=3:1. |
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| 16. |
The graph shows how the magnification m produced by a thin lens varies with image distance v. What is the focal length of the lens used?(A) b²/ac(B) a/c(C) b²c/a(D) b/c |
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| 17. |
Two blocks A and B of masses mA = 1 kg and mB = 3 kg are kept on the table as shown in figure. The coefficient of friction between A and B is 0.2 and between B and the surface of the table is also 0.2. The maximum force F that can be applied on B horizontal, so that the block A does not slide over the block B is:[Take g = 10 m/s²] (A) 8 N (B) 16 N(C) 12 N (D) 40 N |
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Answer» Answer: B. will be CORRECT answers HOPE the answer will help you |
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| 18. |
An NPN transistor is used in common emitter configuration as an amplifier with 1 kΩ load resistance. Signal voltage of 10 mV is applied across the base-emitter. This produces a 3 mA change in the collector current and 15 µA change in the base current of the amplifier. The input resistance and voltage gain are:(A) 0.67 kΩ, 300(B) 0.67 kΩ, 200(C) 0.33 kΩ, 1.5(D) 0.33 kΩ, 300 |
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Answer» Answer: MARK the answer BRAINLIESTS please |
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| 19. |
In a certain oscillatory system, the amplitude ofmotion is 5 m and the time period is 4 s. The timetaken by the particle for passing between pointswhich are at distances of 4 m and 2 m from thecentre and on the same side of it will be(a) 0.30 s (b) 0.32 s (c) 0.33 s (d) 0.35 s |
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Answer» Answer: C)0.33s Explanation: The equation for POSITION of particle at TIME 't' is given by x=(A)sin(ωt) ω=2π/T ,where T is time PERIOD =2π/4 =π/2 4=5sin(πt/2) =>t=(53⁰)2/π=(0.92)(2)/π=0.58s similarly; 2=5sin(πt/2) t=(0.41)(2)/π =0.26s the particle is at 4m and 2m when t₁=0.58s and t₂=0.26s ∴ time interval is given by t₁-t₂=0.3248=0.33 (APPROX) Hope this answer helped you, if so please mark brainliest :) |
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| 20. |
What's the total internal energy of 1 mole of mono atomic gas at temperature T |
Answer» One mole of an ideal MONOATOMIC gas is heated at a constant pressure from 0oC to 100oC. Then the change in the INTERNAL energy of the gas is (GIVEN R=8. 32Jmol−1K−1)..plzzz mark the answer BRAINLIEST! ! ❤✌☺ |
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| 22. |
A circular disc of radius b has a hole of radius a at its centre (see figure). If the mass per unit area of the disc varies as (σ₀/r) then the radius of gyration of the disc about its axis passing through the centre is :(A) (a + b)/3(B) √[(a² + b² + ab)/3](C) (a + b)/2(D)√[(a² + b² + ab)/2] |
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| 23. |
A convex lens (of focal length 20 cm) and a concave mirror, having their principal axes along the same lines, are kept 80 cm apart from each other. The concave mirror is to the right of the convex lens. When an object is kept at a distance of 30 cm to the left o the convex lens, its image remains at the same position even if the concave mirror is removed. The maximum distance of the object for which this concave mirror, by itself would produce a virtual image would be:(A) 20 cm (B) 10 cm(C) 30 cm (D) 20 cm |
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Answer» Answer: The correct answer is d)20 CM because the image produced is half |
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| 24. |
A cell of internal resistance r drives current through an external resistance R. The power delivered by the cell to the external resistance will be maximum when:(A) R = 0.001 r (B) R = 1000 r(C) R = 2r (D) R = r |
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Answer» A cell of internal RESISTANCE r drives current through an external resistance R. The power delivered by the cell to the external resistance will be MAXIMUM when: for step by step explanation refer to attachment correct OPTION is d) R= rfollow me! |
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| 25. |
A massless spring (k = 800 N/m), attached with a mass (500 g) is completely immersed in 1 kg of water. The spring is stretched by 2 cm and released so that it starts vibrating. What would be the order of magnitude of the change in the temperature of water when the vibrations stop completely? (Assume that the water container and spring receive negligible heat and specific heat of mass = 400 J/kg K, specific heat of water = 4184J/kg K)(A) 10⁻³ K (B) 10⁻⁴(C) 10⁻¹ K (D) 10⁻⁵ K |
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Answer» By LAW of CONSERVATION of energy 21 KX2=(m1s1+m2s2)ΔT ΔT=438416×10−2=3.65×10−5. |
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| 26. |
A thin convex lens L (refractive index = 15) is placed on a plane mirror M. When a pin is placed at A, such that OA = 18 cm, its real inverted image is formed at A itself, as shown in figure. When a liquid of refractive index μ₁is put between the lens and the mirror, the pin has to be moved to A’. such that OA’ = 27cm, to get its inverted real image at A’itself. The value of μ₁ will be:(A) √2 (B) 4/3(C) √3 (D) 3/2 |
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| 27. |
A plane is inclined at an angle α = 30º with respect to the horizontal. A particle is projected with a speed u = 2 ms⁻¹, from the base of the plant, making an angle θ = 15º with respect to the plane as shown in the figure. The distance from the base at which the particle hits the plane is close to (Take g = 10 ms²)(A) 18 cm (B) 14 cm(C) 26 cm (D) 20 cm |
Answer» Given:A PLANE is inclined at an angle α = 30º with respect to the horizontal. A particle is projected with a speed u = 2 ms⁻¹, from the base of the plant, making an angle θ = 15º with respect to the plane as shown in the figure. To find:Range of projectile Calculation:Let the angle of projectile w.r.t to ground be Now, please REMEMBER the GENERAL formula for range on inclined plane : PUTTING all the available values: So, FINAL answer: |
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| 28. |
A moving coil galvanometer has a coil with 175 turns and area 1 cm²It uses a torsion band of torsion constant 10⁻⁶ N – m/rad. The coil is placed in a magnetic field B parallel to its plane. The coil deflects by 1⁰ for a current of 1 mA. The value of B (in Tesla) is approximately:-(A) 10⁻³(B) 10⁻¹(C) 10⁻⁴(D) 10⁻² |
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Answer» The moving engagement was devised by Kelvin and later on modified by D' Arsonaval . this is used for detection and measurement of small electric current. Theory :Torque on the coil = NIAB where ,n = number of turns in the coil A = area and B = magnetic FIELD induction of radial magnetic field in which the call is suspended. ↪due to the torque , the coil ROTATES. no restoring torque is developed in the suspension wire. in EQUILIBRIUM, deflecting couple= restoring torque niAB = C∅ Given : i = 1 mA ∅ = 1° = π/180 radian A = 1 cm² n = 175 turns C∅= niAB put the given values correct option a)Follow me! |
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| 29. |
If 'M' is the mass of water that rises in a capillary tube of radius 'r', then mass of water which will rise in a capillary tube of radius ′2r′ is:(A) 4 M(B) M/2(C) M(D) 2 M |
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Answer» If 'M' is the MASS of WATER that rises in a capillary tube of RADIUS 'R', then mass of water which will rise in a capillary tube of radius ′2r′ is: SEE the attachment correct option is d) 2M I hope it helps you! |
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| 30. |
The magnitude of the magnetic field at the centre of an equilateral triangular loop of side 1 m which is carrying a current of 10 A is: [Take μₒ = 4π × 10⁻⁷ NA⁻²] (A) 9μT (B) 1μT(C) 3μT (D) 18μT |
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Answer» Answer: A. will be CORRECT answers Mark the answer BRAINLIESTS please |
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| 31. |
A bullet of mass 20 g has an initial speed of 1 ms⁻¹ just before it starts penetrating a mud wall of thickness 20 cm. If the wall offers a mean resistances of 2.5 10⁻² N, the speed of the bullet after emerging from the other side of the wall is close to(A) 0.7 ms⁻¹ (B) 0.3 ms⁻¹(C) 0.1 ms⁻¹ (D) 0.4 ms⁻¹ |
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Answer» m=20g,u=1m/s,V=? S=20×10−2m a=20×10−3−2.5×10−2m/s2 v2=u2+2as v2=1−2×20×10−32.5×10−2×10020 v=21≃0.7m/s |
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| 32. |
The figure shows a square loop L of side 5 cm which is connected to a network of resistances. The whole set up is moving towards right with a constant speed of 1 cms⁻¹. At some instant, a part of L is in a uniform magnetic field of 1 T, perpendicular to the plane of the loop. If the resistance of L is 1.7 Ω, the current in the loop at that instant will be close to :(A) 115 μA (B) 170 μA(C) 60 μA (D) 150 μA |
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Answer» Answer: B. will be CORRECT answersExplanation: Mark the answer brainliests PLEASE |
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| 33. |
A very long solenoid of radius R is carrying current I(t) = kteᵃᵗ (k > 0), as a function of time (t ≥ 0). counter clockwise current is taken to be positive. A circular conducting coil of radius 2R is placed in the equatorial plane of the solenoid and concentric with the solenoid. The current induced in the outer coil is correctly depicted, as a function of time, by:- |
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Answer» B. is the CORRECT answers when current increase in solenoid then the coil will resist the change and a current induced opposite DIRECTION to the solenoid.but after some current will become constant in the solenoid and rate of change of change will be ZERO that time so induced current will also become zero.Explanation: hope the answer will help you ... MARK the answer brainliests please |
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| 34. |
Group the following into conductor and insulator :aluminium, graphite, rubber, copper, plastic, sulphur, ceramic, silver, iron, wood and paper |
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Answer» Answer: conductors:- ALUMINIUM, graphite, copper, silver, iron, INSULATORS:- RUBBER, plastic, sulphur, ceramic, wood, paper |
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| 35. |
The reflecting surfaces of two mirrors make an angle90° with each other.If a incident of one mirror has an angle of incidence of 30°,draw the reflected from the second mirror.what will be its angle of reflection ? |
Answer» ANSWER:ANGLE of reflection = 60° Explanation :Refer the attached PICTURE. |
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| 36. |
A body is projected with an initial upward velocity component of 80ms and horizontal veocity component of 100ms.find the position and velocity of the body after 3seconds. |
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Answer» Answer: for answer MAKE me BRINLIST and also GIVE me 5 STAR and also give me FFF |
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| 37. |
The value of numerical aperature of the objective lens of a microscope is 1.25. If light of wavelength 5000 Åis used, the minimum separation between two points, to be seen as distinct, will be :(A) 0.48 μm (B) 0.38 μm(C) 0.24 μm (D) 0.12 μm |
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Answer» The minimum separation between two points, to be seen as DISTINCT, will be: (C) 0.24 μm This can be calculated as FOLLOWS:
NA = ( 0.61λ ) / d
= ( 0.61 X 5000 x 10⁻¹⁰ ) / 1.25 = 2.4 x 10 ⁻⁷ m = 0.24 μm |
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| 38. |
A solid sphere and solid cylinder of identical radii approach an incline with the same linear velocity (see figure). Both roll without slipping all throughout. The two climb maximum heights ʰₛₚₕ and ʰcᵧₗ on the incline. The ratio (ʰₛₚₕ)/ (ʰcᵧₗ) is given by:(A) 1 (B) 4/5(C) 2/√5(D) 14/15 |
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Answer» Here is your answer MATE |
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| 39. |
In the circuit shown, a four wire potentiometer is made of a 400 cm long wire, which extends between A and B. The resistance per unit length of the potentiometer wire is r = 0.01 Ω/cm. If an ideal voltmeter is connected as shown with jockey J at 50 cm from end A, the expected reading of the voltmeter will be:(A) 0.75V (B) 0.20V(C)0.25V (D)0.50V |
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Answer» C)0.25V HOPE THIS HELPS PLEASE MARK AS BRAINLIEST |
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| 40. |
A metal wire of resistance 3Ω is elongated to make a uniform wire of double its previous length. The new wire is now bent and the ends joined to make a circle. If two points on this circle make an angle 60º at the centre, the equivalent resistance between these two points will be: (A) (12/5)Ω(B) (5/3)Ω(C) (5/2)Ω(D) (7/2)Ω |
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| 41. |
In an experiment, brass and steel wires of length 1 m each with areas of cross section 1 mm² are used. The wires are connected in series and one end of the combined wire is connected to a rigid support and other end is subjected to elongation. The stress requires to produced a new elongation of 0.2 mm is [Given, the Young’s Modulus for steel and brass are respectively 120 × 10⁹ N/m² and 60 × 10⁹ N/m²](A) 1.8 × 10⁶ N/m²(B) 0.2 × 10⁶ N/m²(C) 1.2 × 10⁶ N/m²(D) 4.0 × 10⁶ N/m² |
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| 42. |
The resistive network shown below is connected to a D.C. source of 16 V. The power consumed by the network is 4 Watt. The value of R is: (A) 8 Ω (B) 6 Ω(B) 16 Ω (D) 1 Ω |
Answer» answer : option (A) 8Ωfirst of all find the equivalent resistance. two 4R resistors are joined in parallel combination. 1/R₁ = 1/4R + 1/4R ⇒R₁ = 2R 6R and 12R are joined in parallel combination. 1/R₂ = 1/6R + 1/12R = 3/12R = 1/4R ⇒R₂ = 4R now R₁ , R₂ , two R's are joined in SERIES combination. Req = R₁ + R₂ + R + R = 2R + 4R + R + R = 8R now use FORMULA, P = V²/Req ⇒4 = (16)²/8R ⇒8R = 256/4 = 64 ⇒R = 8Ω hence option (A) is correct choice. also read similar questions : A body is in pure rotation. The linear SPEED v of a particle, the distance r of the particle from the axis and the angul... The equivalent resistance between points X and Y in the following figure is _____ (A) 4 Ω (B) 2 Ω (C) 1 Ω (D) 3 Ω |
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| 43. |
The resistance of a galvanometer is 50 ohm and the maximum current which can be passed through it is 0.002 A. What resistance must be connected to it in order to convert it into an ammeter of range 0 – 0.5 A?(A) 0.2 ohm (B) 0.002 ohm(C) 0.02 ohm (D) 0.5 ohm |
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Answer» A. will be correct ANSWERS |
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| 44. |
One mole of an ideal gas passes through a process where pressure and volume obey the relation P = Pₒ[1- (Vₒ/V)².1/2]. Here Pₒ and Vₒ are constants. Calculate the change in the temperature of the gas if its volume change from Vₒ to 2Vₒ(A) (1/4)(PₒVₒ/R)(B) (1/2)(PₒVₒ/R)(C) (5/4)(PₒVₒ/R)(D) (3/4)(PₒVₒ/R) |
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| 45. |
Two coils P and Q are separated by some distance. When a current of 3 A flows through coil P a magnetic flux of 10⁻³ Wb passes through Q. No current is passed through Q. When no current passes through P and a current of 2 A passes through Q, the flux through P is:(A) 6.67 × 10⁻³ Wb (B) 6.67 × 10⁻⁴ Wb (C) 3.67 × 10⁻⁴ Wb(D) 3.67 × 10⁻³ Wb |
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| 46. |
If Surface tension (S), Moment of Inertia (I) and Planck’s constant (h), were to be taken as the fundamental units, the dimensional formula for linear momentum would be:(A) S¹ /² I¹ /² h⁰ (B) S¹ /² I³ /² h⁻¹(C) S³ /² I¹ /² h⁰ (D) S¹ /² I¹ /² h⁻¹ |
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Answer» A. will be correct ANSWERS MARK the ANSWER BRAINLIESTS please |
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| 47. |
In the density measurement of a cube, the mass and edge length are measured as (10.00±0.10) kg and (0.10±0.01)m, respectively. The error in the measurement of density is:(A) 0.10 kg/m³ (B) 0.31 kg/m³(C) 0.07 kg/m³ (D) 0.01 kg/m³ |
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Answer» Given : In the density measurement of a cube, the MASS and EDGE length are measured as (10.00 ± 0.10) kg and (0.10 ± 0.01)m, respectively. To find : The error in the measurement of density. solution : we know, volume of cube = (edge length)³ so, density of cube = mass of cube/volume of cube ⇒d = m/V = m/a³ To find error, formula will be ... ∆d/d = ∆m/m + 3∆a/a ......(1) here mass of cube = (10.00 ± 0.10) kg so, m = 10, ∆m = 0.1 edge length of cube = (0.1 ± 0.01) m so, a = 0.1 , ∆a = 0.01 now find error in the density using EQ (1). ∆d/10⁴ = 0.1/10 + 3 × 0.01/0.1 ⇒∆d/d = (0.01 + 3 × 0.1) = 0.31 kg/m³ Therefore the error in the measurement of density is 0.31 kg/m³. Therefore option (B) is correct CHOICE. |
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| 48. |
When heat Q is supplied to a diatomic gas of rigid molecules at constant volume its temperature increases by ∆T. The heat required to produce the same change in temperature, at constant pressure is :(A) (3/2)Q (B) (5/3)Q (C) (7/5)Q (D) (2/3)Q |
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| 49. |
A parallel plate capacitor has 1μF capacitance. One of its two plates is given + 2μC charge and the other plate, +4μC charge. The potential difference developed across the capacitor is:(A) 3V (B) 1V(C) 5V (D) 2V |
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Answer» Answer: HOPE the answer will help you |
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| 50. |
The position of a particle as a function of time t, is given by x(t) = at + bt² - ct³ where a, b and c are constants. When the particle attains zero acceleration, then its velocity will be:(A) a+(b²/4c)(B) a+(b²/c)(C) a+(b²/2c)(D) a+(b²/3c) |
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