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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. |
A wheel intially at rest, starts rotating with a uniform angular acceleration. The wheel rotates through an angle theta_1 , in first second and through a angle theta_2 , in next two second. The value of theta_2 / theta_1 |
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Answer» 8 |
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| 2. |
An object placed at a distance of 10 cm from a convex lens of 20 cm focal length.Find out the position of image. |
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Answer» Solution :Since the object is PLACED between P and F, the image is VIRTUAL. USING 1/v-1/u=1/f Here u=-10cm,f=20cm. THEREFOR 1/`nu |
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| 3. |
Explain resolving power for optical instruments and explain resolving power of telescopes. |
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Answer» Solution :Due to diffraction phenomenon in optic instrument it is difficult to see objects and i images that are too close. he angular resolution of the TELESCOPE determined by the objective of the telescope b increasing the magnification through the ey piece, resolution does not get. Eye piece increas the magnification produced by objective. When a parallel beam of light is incident on convex lens due to this lens the ray beam focus on a point but due to diffraction it will t focus in a finite area instead of being focus : one point. This is shown in figure.  Here due to convex lens dark and bright ring are obtained SIMULTANEOUSLY around the CENTRA bright region. These ring is called Airy.s rings. The radius of the central bright region is if fi the focal length of the lens and 2a is th: diameter of the circular aperture or the diamete of the lens, whichever is smaller then the linea width of central maximum `=(1.22lamdaf)/(2a)` `:.` Radius of central maximum, `r_(0)=(1.22lamdaf)/(2a)=(0.61lamdaf)/(a)` `:.Deltatheta~~(0.61lamda)/(a)` IMPLYING where `Deltatheta` is the minimum angle required to see the two images just separated. Here `Deltatheta` is angular resolution of telescope. An inverse of an angular resolution is called energy resolution or geometric resolution. Thus, when the diameter of the objective is greater, `Deltatheta` will be smaller. That is for a telescope as a is larger then its power of resolution is higher. Resolution power : The ability of an optical instrument to produce distinctly SEPARATE images of two closely placed object is called its resolving power. |
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| 4. |
A body of mass 5 kg rotates at the rate of 5 Hz. if its radius of gyration about the given axis is 0.2m . then its K.E.- |
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Answer» 39.4 J |
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| 5. |
If the earth were to suddenly contract to 1/nth of its present size without any change in its mass, the duration of the new day will be nearly |
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Answer» 24/n hours `I_(1)omega_(1)=I_(2)omega_(2)` `2/5MR^(2)xx(2pi)/24=2/5M(R/n)^(2)xx(2pi)/TrArrT=24/(n^(2))` HOUR |
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| 6. |
Draw the phasor diagram for the devices X. |
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Answer» SOLUTION :Graph of voltage and current with time VARIATION of reactance with frequency (GRAPHICAL variation) Reactance of the CAPACITOR varies in inverse Proportion to the frequency i.e. `X_c alpha (1)/(u) `  
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| 7. |
The Phenomena of photoelectric effect and Compton effect will conform |
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Answer» Wave theory of LIGHT |
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| 8. |
A glass sphere of radius 2R and refractive index n has a spherical cavity of radius R, concentric with it. Q. When viewer is ono left side of the hollow sphere, what will be the shift in position of the object? |
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Answer» `((n+1))/((n-1))R`,right `(1)/(v)-(n)/((-R))=((1-n))/((-2R))` which on solving for v yields `v=-((2R)/(n+1))` Image is on the right of refracting surface S. Shift `=` Real depth `-` Apparent depth `=R-((2R)/(n+1))=((n-1))/((n+1))R` (ii) When the viewer is on the right, TWO REFRACTIONS take place at surface `S_(1)` and `S_(2)`. For refraction at surface `S_(1) :` `(n)/(v_(1))-(1)/((-2R))=((n-1))/((-R))` which ono solving for `v_(1)` yields `v_(1)=-(2nR)/(2n-1)` The first lies to the left of `S_(1)` act as object for refraction at the second surface. We have to shift the origin of Cartesian coordinate system to the VERTEX of `S_(2)`. The object distance for the second surface is `u_(2)=-[(2nR)/(2n-1)+R]=-((4n-1)/(2n-1))R` `(1)/(v_(2))=-(n)/([(4n-1)/(2n-1)])=(1-n)/(-2R)` On solving for `v_(2)` , we get `v_(2)=-(2(4n-1))/(3(n-1))R` The minus sign shows that image is virtual and lies to the left of `S_(2)`. Shift `=` Real depth `-` Apparent depth `=3R-(2(4n-1)R)/((3n-1))=((n-1))/((3n-1))R` |
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| 9. |
A glass sphere of radius 2R and refractive index n has a spherical cavity of radius R, concentric with it. Q. When viewer is on right side of the hollow sphere, what will be apparent change in position of the object? |
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Answer» `((n-1))/((3n+1))R`,towards left `(1)/(v)-(n)/((-R))=((1-n))/((-2R))` which on SOLVING for v YIELDS `v=-((2R)/(n+1))` Image is on the right of refracting surface S. Shift `=` REAL depth `-` Apparent depth `=R-((2R)/(n+1))=((n-1))/((n+1))R` (ii) When the viewer is on the right, two refractions take place at surface `S_(1)` and `S_(2)`. For refraction at surface `S_(1) :` `(n)/(v_(1))-(1)/((-2R))=((n-1))/((-R))` which ono solving for `v_(1)` yields `v_(1)=-(2nR)/(2n-1)` The first lies to the left of `S_(1)` act as object for refraction at the second surface. We have to shift the origin of Cartesian coordinate system to the VERTEX of `S_(2)`. The object distance for the second surface is `u_(2)=-[(2nR)/(2n-1)+R]=-((4n-1)/(2n-1))R` `(1)/(v_(2))=-(n)/([(4n-1)/(2n-1)])=(1-n)/(-2R)` On solving for `v_(2)` , we get `v_(2)=-(2(4n-1))/(3(n-1))R` The minus sign shows that image is virtual and lies to the left of `S_(2)`. Shift `=` Real depth `-` Apparent depth `=3R-(2(4n-1)R)/((3n-1))=((n-1))/((3n-1))R` |
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| 10. |
A particle starts from rest in circular path of radius R = 2 m such that its angular velocity is omega = (pi t)/(3) rad/s. Find the magnitude of average velocity of particle when it has maoved by angle 60^(@) from its initial position. |
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Answer» `sqrt3 m//s` `omegta = (PI t)/(3) ` `alpha = ( d omega)/(dt) = (pi)/(3) rad //s^(2) ` `theta = (pi)/(3) =1/2 alpha t^(2)` `pi/3 =1/2 (pi//3) t^(2) ` `t = sqrt2s` Now displacement of paraticle is given as `d= R =2M` `V = (2)/(sqrt2) = 2sqrtm//s` 
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| 11. |
AB is the semi-circular object of diameter (f/2) where f is the focal length, placed on the principalaxis of a concave mirror with point A at the centre of curvature. A'B' is the image formed. The correct option is : (Assume diameter of the object is along principal axis) |
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Answer»
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| 12. |
Large transformer when used for sometime becomevery hot andcooled by circulating oil the heating of the transformers is due to |
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Answer» the HEATING effect of the current alone |
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| 13. |
The resultant of two vectors A and B is perpendicular to the vector A and its magnitude is equal to half the magnitude of vector B. The angle between A and B is - |
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Answer» ` cos theta=-1/2` ` theta = 120^@` |
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| 14. |
The velocity of the small ball of mass M and density d_(1) when dropped in a container filled with glycerine becomes constant after sometimes. If the density of glycerine is d_(2), the viscous force acting on the ball is : |
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Answer» `MG(1-d_(2)//d_(1))` |
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| 15. |
किसी वस्तु के द्वारा तय की गई कुल दूरी 5 की प्रथम एक तिहाई, द्वितीय एक तिहाई तथा तृतीय एक तिहाई दूरी में चालें क्रमशः V, 2V और 3v हैं। इसकी औसत चाल होगी |
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Answer» V |
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| 16. |
A point charge Q is placed at the centre of an uncharged conducting shell. Let r be the distance of a point from Q. The point may lie either inside or outside the shell. The electric intensity at the point will be Q//4piepsilon_(0)r^(2) if the pont lies |
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Answer» INSIDE the shell but not OUTSIDE it |
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| 17. |
The number of hydrogen bonds between guanine and cytosine, and between adenine and thymine in DNA is |
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Answer» 1,2 i.e., `G_=C and A=T` |
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| 18. |
A projectile is launched from the surface of the earth with a very high speed u at an angle theta with vertical . What is its velocity when it is at the farthest distance from the earth surface. Given that the maximum height reached when it is launched vertically from the earth with a velocity v = sqrt((GM)/(R)) |
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Answer» `(U "cos" theta)/(2)` |
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| 19. |
A hollow cylinder (rho=2.2xx10^(-8)Omega-m) of length 3 m has inner and outer diameters are 2 min and 4 mm respectively. The resistance of the cylinder is |
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Answer» `0.35xx10^(-3)OMEGA` |
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| 20. |
The human eye resembles the _____. |
| Answer» SOLUTION :PHOTOGRAPHIC CAMERA | |
| 21. |
What is electric dipole?Why do we assume the electric dipole to be small in size in most of the cases? |
| Answer» Solution :A SYSTEM of two equal and opposite charges placed at some separation is called ELECTRIC dipole. The concept of electric dipole is USEFUL at a molecular level and that is why we refor small dipole as an ideal dipole. In some molecules, the centre of positive charge gets shifted from the centre of NEGATIVE charge. Such a molecule BEHAVES like one electric dipole. | |
| 22. |
A man of weight mg is moving upwards in a rocket with acceleration of 4 g. His apparent weight inside the rocket will be: |
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Answer» Zero So correct CHOISE is (C). |
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| 23. |
The pole pieces of the magnet used in a pivoted coil galvanometer are |
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Answer» PLANE SURFACES of a BAR magnet |
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| 24. |
Assertion : In a cavity within a conductor, the electric field is zero. Reason : Charges in a conductor reside only at its surface. |
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Answer» If both assertion and reason are TRUE and reason is the correct explanation of assertion Both the assertion and reason are true and the reason is the correct explanation. It is because the charges are only at the surface of a CONDUCTOR, the charge enclosed in the Gaussian surface in the CAVITY is zero. The field is THEREFORE zero. |
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| 25. |
A maximum current of 0.5 mA can be passed through a galvanometer of resistance 20Omega. Calculate the resistance to be connected in series to convert it into a voltmeter of range (0- 5)V. |
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Answer» Solution :R = G (n -1), where `n=V/(V_(g))` `V=5V,V_(g)=i_(g)G=0.5xx10^(-3)xx20=10^(-2)V` `thereforen=500` and `R=20(500-1)=9980Omega` |
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| 26. |
A radioactive sample can decay by two different processes. The half-life for the first process is T_1and that for the second process is T_2Find the effective half - life T of the radioactive sample. |
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Answer» Solution :Let N be the total number of ATOMS of the radioactive sample initially. Let `(dN_1)/(dt)` and `(dN_2)/(dt)` be the initial rates of DISINTEGRATIONS of the radioactive sample by the two processes respectively. Then `(dN_1)/(dt)=lambda_1N` and `(dN_2)/(dt)=lambda_2N` where `lambda_1` and `lambda_2` are the decay constants for the first and second processes respectively. The initial rate of disintegrations of the radioactive sample by both the processes `=(dN_1)/(dt)+(dN_2)/(dt)=lambda_1N+lambda_2N=(lambda_1+lambda_2)N` If `lambda`is the effective decay CONSTANT of the radioactive sample, its initial rate of disintegration. `(dN)/(dt)=lambdaN`, But `(dN)/(dt)=(dN_1)/(dt)+(dN_2)/(dt)` `lambdaN=(lambda_1+lambda_2)N, lambda=lambda_1+lambda_2` `0.693/T_1+0.693/T_2=0.693/T , 1/T=1/T_1+1/T_2 ,T=(T_1T_2)/(T_1+T_2)` |
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| 27. |
The color code of a carbon resistor is Brown -Red - Brown -Gold. What is its resistance ? |
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Answer» Solution :`12XX10 PM 5%` `(120 pm 5%)Omega` |
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| 28. |
निम्न में से राष्ट्रीय संसाधन है? |
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Answer» श्मशान |
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| 29. |
कोयला किस प्रकार का संसाधन है? |
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Answer» अनवीकरणीय |
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| 30. |
A crown glass prism with refracting angle 5° is to be achromatised for red and blue light with flint glass prism. Angle of the flint glass prism should be ( Given for grown glass mu_(r)-1.513, mu_(b)= 1.523, for flint glass mu_(r)=1.645,mu_(b)=1.665) |
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Answer» 1.5° |
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| 31. |
The electric and magnetic field in an electromagnetic wave are related as __________. |
| Answer» SOLUTION :`B_(0)=(E_(0))/(C )` | |
| 32. |
Null points are obtained on equatorial line, when the north pole of the bar magnet is pointing towards geographical north. Why ? |
| Answer» SOLUTION :The LINES of earth.s magnetic FIELD meet those of the magnet in the opposite direction only on equatorial LINE. | |
| 33. |
A hoop of radius r and mass m rotating with an angular velocity omega_(0) is placed on a rough horizontal surface. The initial velocity of the centre of the hoop is zero. What will be the velocity of the centre of the loop when it ceases to slip? |
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Answer» `(romega_(0))/(3)` `therefore omega=(omega_(0))/(2)` `therefore V_(CM)=(omega_(0)r)/(2)` |
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| 34. |
Match the following{:("List - I ","List - II "),("a)Silver lining of mountains " ,"e)polarization by refraction "),("b) Rectilinear propagation light ","f) transverse nature of light "),("c) Polarization","g)diffraction "),("d) Pile of plates","h) ray optic "):} |
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Answer» `a - H, b - G, c - F, d - E` |
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| 35. |
A reversible engine converts one third input into work. When the temperature of the sink is reduced by 62 °C, then efficiency of the engme 1s tripled. The temperature of the source and sink are |
| Answer» Answer :C | |
| 36. |
Consider the following compounds I to IV with respect to their S_(N)2 reacrtivity with a given nucelophile{:(CH_(3)CH_(2)Br,CH_(3)CH_(2)CH_(2)Br),(""I,""II),((CH_(3))_(2)CHCH_(2)Br,(CH_(3))_(3)C CH_(2)Br),(""III,""IV):} |
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Answer» `IV GT III gt II gt I` `(i)gt(ii)gt(iii)gt(iv)` |
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| 37. |
Two identical thin plano-convex glass lenses (refractive index 1.5) each having radius of curvature of 20 cm are placed with their convex surface in contact at the centre. The intervening space is filled with oil of refractive index 1.7. The focal length of the combination is : |
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Answer» `-20` cm `1/f=(n-1)((1)/(R_1)-(1)/(R_2))` `(1)/(f_1)=(1.5-1)((1)/(infty)-(1)/(-20))=0.5xx(1)/(20)=(1)/(40)` and `(1)/(f_2)=(1.7-1)((1)/(-20)-(1)/(+20))=-0.7xx(2)/(20)=-(7)/(100)` `(1)/(f_3)=(1.5-1)((1)/(20)-(1)/(-infty))=0.5xx(1)/(20)=(1)/(40)` `THEREFORE` Focal length of combination, `1/f=(1)/(f_1)+(1)/(f_2)+(1)/(f_3)`  `1/f=(1)/(40)-(7)/(100)+(1)/(40)` `=(2.5-7+2.5)/(100)` `therefore 1/f=-(2)/(100)` `therefore f=-50` cm |
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| 38. |
Paragraph : A particle vibrates in S.H.M. along a straight line. Its velocity is 4 cm/s when its displacement is 3 cm and velocity is 3 cm/s when displacement is 4 cm. Starting from origin a body oscillates simple harmonically with a period of 2 s. After what time will its kineti energy be 75% of the total energy. |
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Answer» `(1)/(6)s` `implies(1)/(2)m OMEGA^(2)(R^(2)-y^(2))=(3)/(4)((1)/(2)m omega^(2)r^(2))` `r^(2)-y^(2)=(3)/(4)r^(2)""impliesy=r//2` Now `y=r sin omegatimplies( r )/(2)=r sin" wt"impliessin omegat=(1)/(2)` `implies""omegat=pi//6""implies(2pi)/(T)t=(pi)/(6)""implies""t=(T)/(12)` So CORRECT choice is (a). |
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| 39. |
Paragraph : A particle vibrates in S.H.M. along a straight line. Its velocity is 4 cm/s when its displacement is 3 cm and velocity is 3 cm/s when displacement is 4 cm. A particle of mass m executes simple harmonic motion with amplitude 'a' and frequency v. The average kinetic energy during its motion from the position of equilibrium to the end is : |
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Answer» `(1)/(4)ma^(2)v^(2)` `=(1)/(4)m(2pi v)^(2).a^(2)=pi ma^(2)v^(2)` So correct CHOICE is (d). |
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| 40. |
Paragraph : A particle vibrates in S.H.M. along a straight line. Its velocity is 4 cm/s when its displacement is 3 cm and velocity is 3 cm/s when displacement is 4 cm. The displacement of an object attached to a spring and executing simple harmonic motion is given by x=2xx10^(-2)cos pi t metres. The time at which the maximum speed first occurs is : |
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Answer» 0.75 s Comparing it with standard from `x=r cos omegat`. We have `omega=piimplies(2pi)/(T)=pi` `IMPLIES""T=2` second Now in time `t=(T)/(4)` the particle GOES from extreme POSITION to MEAN position where K.E. becomes maximum. So, correct CHOICE is (d). |
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| 41. |
Paragraph : A particle vibrates in S.H.M. along a straight line. Its velocity is 4 cm/s when its displacement is 3 cm and velocity is 3 cm/s when displacement is 4 cm. The displacement of a particle varies according to the relation X = 4 (cos pi t+sin pi t). The amplitude of the particle is : |
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Answer» `-4` since `cos pi t` and `sin pi t` are out of phase by `90^(@)`. `:.` Amplitude `A=sqrt(4^(2)+4^(2))=4sqrt(2)` Hence CORRECT choice is ( C ). |
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| 42. |
Paragraph : A particle vibrates in S.H.M. along a straight line. Its velocity is 4 cm/s when its displacement is 3 cm and velocity is 3 cm/s when displacement is 4 cm. Two simple harmonic motions are represented by the equations y_(1)=0.15m(100pit+pi//3) and y = 0.1 cos pit. The phase difference of the velocity of particle 1 with respect to the velocity of particle 2 is : |
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Answer» `-PI//6` `v_(2)=(dy_(2))/(dt)=-0.1xx pi SIN pi t=0.1xx pi cos (pi t+pi//2)` PHASE DIFF. `Delta phi_(12)=(pi)/(3)-(pi)/(2)=-(pi)/(6)` Thus correct choice is (a). |
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| 43. |
Paragraph : A particle vibrates in S.H.M. along a straight line. Its velocity is 4 cm/s when its displacement is 3 cm and velocity is 3 cm/s when displacement is 4 cm. A particle executes simple harmonic motion between x=-A and x=+A. The time taken for it to go from 0 to A/2 is T_(1) and go from A/2 to A is T_(2). Then : |
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Answer» `T_(1)lt T_(2)` `(A)/(2)=A sin OMEGA T_(1)""implies""omega T_(1)=(PI)/(6)""implies""(2pi T_(1))/(T)=(pi)/(6)` `T_(1)=(T)/(12)` `:.` Now `T_(1)+T_(2)=(T)/(4)impliesT_(2)=(T)/(4)-(T)/(12)=(T)/(6)` CLEARLY `T_(1) lt T_(2)`. Correct choice is (a). |
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| 44. |
Paragraph : A particle vibrates in S.H.M. along a straight line. Its velocity is 4 cm/s when its displacement is 3 cm and velocity is 3 cm/s when displacement is 4 cm. The length of a simple pendulum executing simple harmonic motion is increased by 21%. The percentage increase in the time period of the pendulum of increased length is : |
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Answer» 0.11 `(T.)/(T)=sqrt(1.21)implies(T.)/(T)=1.1` `implies""T.=1.1T` `:. DELTA T=0.1" T"` %increase in time PERIOD `=(DELTAT)/(T)xx100=(0.1T)/(T)xx100=10%` Hence correct choice is (d). |
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| 45. |
A transparent cube of 15 cm edge contains a small air bubble. Its apparent depth when viewed through one face is 6 cm and when viewed through the opposite face is 4 cm. Then the refractive index of the material of the cube is |
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Answer» 2 |
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| 46. |
The unit of magnetic induction B in SI is : |
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Answer» A/m |
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| 47. |
At a location, the horizontal component of the earth's magnetic field is 0.3 G in the magnetic meridian and the dip angle is 60^@ . The earth's magnetic field at this location in G is |
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Answer» 0.3 `0.3 = B cos 60^@` ` = 0.3 = B xx 1/2 " or " B= 0.6G` |
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| 48. |
Paragraph : A particle vibrates in S.H.M. along a straight line. Its velocity is 4 cm/s when its displacement is 3 cm and velocity is 3 cm/s when displacement is 4 cm. The period of oscillation in seconds is : |
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Answer» SOLUTION :`T=2pi sqrt((x_(2)^(2)-x_(1)^(2))/(u_(1)^(2)-u_(2)^(2)))=2pi sqrt((16-9)/(16-9))=2pi` So correct choice is (b). |
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| 49. |
Paragraph : A particle vibrates in S.H.M. along a straight line. Its velocity is 4 cm/s when its displacement is 3 cm and velocity is 3 cm/s when displacement is 4 cm. The maximum value of acceleration is : |
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Answer» `2" cm"//"s"^(2)` `=5" cm"//"s"^(2)` So correct CHOICE is (d). |
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| 50. |
Paragraph : A particle vibrates in S.H.M. along a straight line. Its velocity is 4 cm/s when its displacement is 3 cm and velocity is 3 cm/s when displacement is 4 cm. If mass of body is 50 g the calculate the total energy of oscillation : |
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Answer» `6.25xx10^(-5)` J `E=(1)/(2)m omega^(2)r^(2)` `=(1)/(2)xx(50xx10^(-3))xx((2PI)/(2pi))^(2)xx(5xx10^(-2))^(2)` `=6.25xx10^(-5)" J"` So correct choice is (a). |
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