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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) : At frequencies above 40 MHz, communication is essentially limited to line of sightpath. (R) : At these frequencies, the antenna is relatively smaller and can be placed at heights of many of wavelengths above the ground. |
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Answer» Both 'A' and 'R' are true and 'R' is the CORRECT explanation of 'A'. |
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| 2. |
Find the phase difference between voltage and current. |
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Answer» |
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| 3. |
निम्न में से कौन सजीव व निर्जीव दोनों के लक्षण प्रदर्शित करता है |
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Answer» जीवाणु |
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| 4. |
A potential difference is applied between the ends of a conductor of uniform radius 1.5 mm. It results in an electric field of strength 250 V/m along the length of conductor. Find the resistivity of the material if current in the conductor is 2A. |
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Answer» `8.83xx10^(-4)Omega-m` |
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| 5. |
एक सजीव, एक निर्जीव की संरचना से साधारणतया किस आधार पर अलग होगा |
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Answer» प्रजनन |
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| 6. |
The refractive indices of water and glass are 1.2 and 1.5 respectively. What will be the refractive index of glass with respect to water ? |
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Answer» 1.6 `n_(gw)=(n_g)/(n_w) therefore m_(gw)=(1.5)/(1.2)=1.25` |
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| 7. |
S is a solid neutral conducting sphere. A point charge q of 1 xx 10^(-6)C is placed at point, C is the center of sphere, and AB is a tangent. BC = 3 m and B = 4m. |
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Answer» The ELECTRIC potentialof the CONDUCTOR is 1.8 kV. Distance `AC` is GIVEN by `AC=5m` Potential at due to charge `q` at `A` is given by `V=(kq)/(AC)=(9xx10^(9)xx1xx10^(-6))/(5)=1.8xx10^(3)=1.8kV` Potential at `B` is given by `V_(B)=(V_(B))_("due to" q)+(V_(B))_("induced")` `{(V)(B))_(i)=` Potential at `B` due to induced charge} THUS `1.8xx10^(3)=(kq)/(AB)+(V_(B))_(i)=2.25xx10^(3)+(V_(B))_(i)` `(V_(B))_(i)=-0.45kV` So (a) and (c) are correct 
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| 8. |
When a capacitor of small capacitance is connected in series withL-R circuit. The alternating current in the circuit increases. Explain why ? |
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Answer» Solution :Addition of capacitor in the GIVEN CIRCUIT decreases the impedance Z of the circuit and hence INCREASES current I in the circuit as `I = (V)/(Z)` where `Z = sqrt(R^(2) + X_(L)^(2))` WITHOUT capacitor and new `Z = sqrt(R^(2) + (X_(L) - X_(C))^(2))` with capacitor. |
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| 9. |
Read the pasage given below as well as theadjoining energylevel diagram and then answer the questions given afterthe passage . As per Bohr'stheory oghydrogen atom the energy of an atom in a statecorresponding to principal quantum numbern is given as : E_(n) = - (13.6)/(n^(2)) eV Theenergyof an atom is the least when itselectron is revolvingin an orbit closets to thenucleus i.e.,theone forwhich n =1. This stateis called the groundstate. When a hydrogen atom receives energyby processes suchas electron collision, theatom may acquire sufficientenergyto raise theelectron to higherenergystates and the atom is then saidto bein an excited state . From theseexcited states the electron can then fall back to a state of lower energy , emitting a photon in the process. The energy level diagram for thestationary states of a hydrogen atom , compouted from Bohr'srelation for energy, is givenin Fig.12.02. Theprincipalquantum number n labels the stationary states in the ascendingorderof energy . Obviously , thehighestenergycorrespondsto n = ooandhasan energy of 0 eV. Thisis the energy of the atom when the electron is completely removed (r= oo) from the nucleus and is at rest. Now answer the follwoingquestions: How can a hydrogenatom receiveenergy to raisethe electron to higherenergy states ? |
| Answer» Solution :At ROOM temperaturea hydrogen atomreceives energyby PROCESSES suchas ELECTRON collisions so as to raisethe elctron to HIGHER energystates. | |
| 11. |
A long straight cable of length l is placed symmetrically along z-axis and has radius a(ltlt l). The cable consists of a thin wire and a co- axial conducting tube. An alternating current I(t) = I_(0) " sin " (2pi vt).Flows down the central thin wire and returns along theco-axial conducting tube. the induced electric at a distance s from thewire inside the cable is E(s ,t) =mu_(0) I_(0) v " cos "(2pivt) In ((s)/(a)) hatk. (i) Calculate the displacement current density inside the cable. (ii) Integrate the displacement current density across the cross- section of the cable to find the total displacement current I^(d). (iii) compare theconduction current I_(0) with thedisplacement current I_(0)^(d). |
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Answer» `(2pi)/(LAMDA^(2))I_(0)LN((a)/(s))sin(2pi upsilont)hat(k)` `=epsilon_(0)mu_(0)I_(0)upsilon(del)/(delt)upsiloncos(2piupsilont)ln.((s)/(a))hatk` `=(1)/(c^(2))I_(0)2piupsilon^(2)(-sin(2piupsilont))ln.((s)/(a))hatk` `=((upsilon)/(c))^(2)2piI_(0)sin(2piupsilont)ln.((a)/(s))hatk=(2pi)/(lambda^(2))I_(0)ln.((a)/(s))sin(2piupsilont)hatk` |
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| 12. |
Two masses m_(1) and m_(2) are connected by means of a light string, that passes over a light pulley as shown in the figure. If m_(1)=2kg and m_(2)=5kg and a vertical force F is applied on the pulley, then find the acceleration of the masses and that of the pulley when (a) f = 35N, (b) F = 70N, (c) F = 140N. |
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Answer» SOLUTION :Since string is massless and friction is ABSENT, hence tension in the string is same every where. (a) Let acceleration of the pulley be `a_(p)`. For `a_(p)` to be non-zero, F.B.D. of the pulley `T gt m_(1)g ""...(1)` `rArr T gt 2g "" ...(2)` From (1) and (2), we get `F gt 2 xx (2g) rArr F gt 40N` Therefore, when `F=35N` `a_(p)=0` and hence `a_(1)=a_(2)=0` As mass of the pulley is negligible `F-2T=0` `rArr T=F//2 rArr T=35N` To lift `m_(2)`, `T ge m_(2)g rArr T ge 50N` Therefore, block `m_(2)` will not move `rArr T-m_(1)g=m_(1)a_(1) rArr 15=2a_(1)` `rArr a_(1)=(15)/(2)m//s^(2)` CONSTRAINT equation `y_(p)+y_(p)-y_(1)=` constant `rArr 2y_(p)-y_(1)=c rArr 2(d^(2)y_(p))/(dt^(2))-(d^(2)y_(1))/(dt^(2))=0` `rArr a_(p)=(a_(1))/(2)=(15)/(4) m//s^(2)` (c) When F = 140N T = 70N `rArr T-m_(1)g=m_(1)a_(1)"" ...(1)` `rArr 70N-20N=2xxa_(1) rArr a_(1)=25m//s^(2)` `T-m_(2)g=m_(2)a_(2)"" ...(2)` `rArr 70N-50N=5a_(2) rArr a_(2)=4m//s^(2)` Constraint equation `y_(p)-y_(2)+y_(p)-y_(1)=c` `rArr 2y_(p)-y_(1)-y_(2)=c` `rArr 2(d^(2)y_(p))/(dt^(2))-(d^(2)y_(1))/(dt^(2))-(d^(2)y_(2))/(dt^(2))=0` `rArr a_(p)=(a_(1)+a_(2))/(2)=(29)/(2)m//s^(2)` 
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| 13. |
A person, sunbathing on a warm day, is lying horizontally on the deck of a boat. Her mass is 59 kg, and the coefficient of static friction between the deck and her is 0.70. Assume that the person is moving horizontally, and that the static frictional force is the only force acting on her in this direction. What is the magnitude of the static frictional force when the boat moves with a constant velocity of +8.0 m/s? |
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Answer» 94N |
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| 14. |
A circular coil and a bar magnet placed nearly are made to move in same direction. If the coil covers a distance of 1 m in 0.5sec and the magnet at a distance of 2m in 1 sec. The induced e.m.f. produced in the coil is: |
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Answer» Zero |
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| 15. |
Which of the following are broad spectrum antibiotics? |
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Answer» I, II only Pencilin-G is a narrow spectrum antibiotics. HENCE OPTION ( C) is correct |
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| 16. |
The potential difference developed across the junction due to migration of majority carriers is called____. |
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Answer» POTENTIAL barrier |
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| 17. |
Assertion: Three equal charges are situated on a circle of radius r such that they form an equilateral triangle, then the electric field intensity at the centre is zero. Reason:The force on unit positive charge at the centroid of an equilateral triangle due to the three equal charges at the three vertices are represented by the three sides of a triangle taken in the same order. (Therefore, electric field intensity at centre is zero.) |
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Answer» Both ASSERTION and REASON are true and Reason is the correct explanation of Assertion |
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| 18. |
The maximum energy in thermal radiation for a source occurs at wavelength 11xx10^(-5)cm. The temperature of the source is n times the temperature of another source of which the wavelength at maximum energy emission is 5.5xx10^(-5) cm The value of n is : |
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Answer» Solution :Since `lamdamT=b`. `:.""lamda_(1)mT_(1)=lamda_(2)mT_(2)` `rArr11xx10^(-5)xxT_(1)=T_(2)XX5*5xx10^(-5)`. Since `T_(1)=nT_(2)`. `:.""11xxnT_(2)=5*5T_(2)`. `rArr""n=1/2` Hence correct choice is (c ). |
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| 19. |
Read the pasage given below as well as theadjoining energylevel diagram and then answer the questions given afterthe passage . As per Bohr'stheory oghydrogen atom the energy of an atom in a statecorresponding to principal quantum numbern is given as : E_(n) = - (13.6)/(n^(2)) eV Theenergyof an atom is the least when itselectron is revolvingin an orbit closets to thenucleus i.e.,theone forwhich n =1. This stateis called the groundstate. When a hydrogen atom receives energyby processes suchas electron collision, theatom may acquire sufficientenergyto raise theelectron to higherenergystates and the atom is then saidto bein an excited state . From theseexcited states the electron can then fall back to a state of lower energy , emitting a photon in the process. The energy level diagram for thestationary states of a hydrogen atom , compouted from Bohr'srelation for energy, is givenin Fig.12.02. Theprincipalquantum number n labels the stationary states in the ascendingorderof energy . Obviously , thehighestenergycorrespondsto n = ooandhasan energy of 0 eV. Thisis the energy of the atom when the electron is completely removed (r= oo) from the nucleus and is at rest. Now answer the follwoingquestions: What does negative value of atomic energy signify? |
| Answer» Solution :Thenegative sing of THEENERGY of anelectron movingin an orbitin HYDROGEN atom mean thatthe electron is boundwiththe NUCLEUS . | |
| 20. |
Explain the theory of interference of light. |
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Answer» Solution :Let `Y_(1)=asinomegat` and `Y_(2)=bsin(omegat+phi)` be two light waves interfere GIVING rise to RESULTANT `Y=Y_(1)+Y_(2)` NET displacement from principle of superposition `Y=Y_(1)+Y_(2)` `Y=asinomegat+bsin(omegat+phi)=asinomegat+b(sinomegatcosphi+cosomegatsinphi)` `=sinomegat(a+bcosphi)+bsinphicosomegat` Let `a+bcosphi=Rcostheta` and `bsinphi=Rsintheta` `Y=Rsinomegatcostheta+Rcosomegatsintheta=Rsin(omegat+theta)` `R=sqrt(a^(2)+b^(2)+2abcostheta)` OR the wave with equal amplitude must be considered. `R=sqrt(2a^(2)+2B^(2)+2acostheta)=2acos.(theta)/(2)` |
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| 21. |
There is a detective submarine installed inside sea water post 26/11 incident to detect terrorists. It is moving with constant speed V_theta along a straight line and it sends a wave which travels with speed V_w = 1100m/sec in water. Initially waves are getting reflected from a fixed Island and the frequency detected by the submarine is found to be 20% more than the original frequency. When a terrorist ship moving towards the submarine with constant speed V_s comes in between the submarine and the island. Frequency of waves reflected from the ship is 80% more than the original frequency.Speed of enemy ship V_sis |
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Answer» 220 m/sec |
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| 22. |
A rod of length 10 cm lies along the principal axis of a concave mirror of focal length 10 cm in such a way that its end closer to the pole is 20 cm away from the mirror. The length of the image is, |
| Answer» SOLUTION :5 cm | |
| 23. |
There is a detective submarine installed inside sea water post 26/11 incident to detect terrorists. It is moving with constant speed V_theta along a straight line and it sends a wave which travels with speed V_w = 1100m/sec in water. Initially waves are getting reflected from a fixed Island and the frequency detected by the submarine is found to be 20% more than the original frequency. When a terrorist ship moving towards the submarine with constant speed V_s comes in between the submarine and the island. Frequency of waves reflected from the ship is 80% more than the original frequency.Bulk modulus of sea water is |
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Answer» `10^9 N//m^2` |
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| 24. |
How does the self inductance of a coil change when an iron rod is introduced into it? Justify your answer in each case. |
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Answer» Solution :Self-inductance of a coil, L= `mu_0n^2lA` On introducing the iron ROD into the coil, the self inductance will become L = `mu_0mu_rn^2lA` where `MU`, is the RELATIVE permeability of iron. since `mu_r`>1, self inductance will INCREASE. |
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| 25. |
The equation of a waves is y = sin (2pi rt//lambda). When it is reflected to at a rigid support its amplitude reduces by 10%. The equation of the replaced wave |
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Answer» `y = 0.9 A sin (2 pi R t//lambda)` |
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| 26. |
A spherical shell of radius R is given charge 3q on its surface and a point charge q is placed at distance R//2 from its centre C. Also there is a charge 2q placed outside the shell at a distance of 2R as shown. Then |
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Answer» The magnitude of electric field at the centre `C` due to CHARGE on the outer SURFACE of shell is `(KQ)/(2R^(2))` before closing the switch `S` |
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| 27. |
There is a detective submarine installed inside sea water post 26/11 incident to detect terrorists. It is moving with constant speed V_theta along a straight line and it sends a wave which travels with speed V_w = 1100m/sec in water. Initially waves are getting reflected from a fixed Island and the frequency detected by the submarine is found to be 20% more than the original frequency. When a terrorist ship moving towards the submarine with constant speed V_s comes in between the submarine and the island. Frequency of waves reflected from the ship is 80% more than the original frequency. Velocity V_0 is |
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Answer» 50 m/sec |
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| 28. |
A capacitor of capacitance 10 muF is charged to 100 V. It is now connected to uncharged capacitor in parallel so that the common potential becomes 40 V. The capacitance of the second capacitor is |
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Answer» `10 mu F` |
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| 29. |
Atypical Angiosperm anther is |
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Answer» Monolobed,MONOTHECOUS and bisporangiate |
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| 30. |
If one of the two electrons of a H_2 molecule is removed, we get a hydrogen molecular ion H_2^(+).In the ground state of an H_2^(+), the two protons are separated by roughly 1.5 Å, and the electron is roughly 1 Å from each proton. Determine the potential energy of the system. Specify your choice of the zero of potential energy. |
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Answer» SOLUTION :If zero of the potential energy is taken at infinity, then in PRESENT set up the potential energy of the system should be `U = 1/(4pi epsi_0)[(q_1q_2)/r_12 + (q_1q_3)/r_13+ (q_2q_3)/(r_23)]` `= 9 XX 10^9 [((+1.60xx10^(-19))(1.60 xx 10^(-19)))/(1.5 xx 10^(-10))+((1.60 xx 10^(-19))(-1.60 xx 10^(-19)))/(1 xx 10^(-10))+((1.60 xx 10^(-19))(-1.60xx10^(-19)))/(1 xx 10^(-10))]J` `=3.072 xx 10^(-18)J =-(3.072 xx 10^(-18))/(1.60 xx 10^(-19)) eV = -19.2eV`. 
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| 31. |
A stretched wire under tension of 64 N vibrating in its fundamental mode is in resonance with a vibrating tuning fork. The vibrating portion of the sonometer wire has a length of 10 cm and mass lg. The vibrating tuning fork in now moved away from the vibrating wire at a constant speed and an observer standing near the sonometer hears one beat per sec. Calculate the speed with which the tuning fork is moved, if the speed of sound in air is 300 m/s. |
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Answer» Solution :As the frequency of a vibrating string is `f=(1)/(2l) sqrt((T)/(mu))` On SUBSTITUTION gives `f=400 Hz`. When the tuning fork is moved away from the observer standing near the sonometer at a constant speed u the apparent frequency of tuning fork will be `f_(R)=f[(V)/(V+u)]` As `f_(R)` is producing beats with f, `f_(R)` is nearly EQUAL to f, i.e., `u lt lt V` so that, `f_(R)=f[1+(u)/(V)]^(-1)=f[1-(u)/(V)]` So beat frequency `DELTAF=f-f_(R)=f|(u)/(V)]` and substituting gives data, `u=V|(Deltaf)/(f)|=300[(1)/(400)]=0.75m//s` |
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| 32. |
When the voltage applied to an X-ray tube increased from V_(1)=15.5kV to V_(2)=31kV the wavelength interval between the K_(alpha) line and the cut-off wavelength of te continuous X-ray spectrum increases by a factor of 1.3. If te atomic number of the element of the target is z. Then the value of (z)/(13) will be: (take hc=1240eVnm and R=1xx(10^(7))/(m)) |
| Answer» Answer :A | |
| 33. |
The magnifying power of a telescope is m. If the focal length of the eye-piece is halved, then its magnifying power is ......... |
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Answer» `(1)/(2m)` If we take `f._e=(f_e)/(2)` `m.=(f_0)/(f_e)=(2f_0)/(f_e)=2(m)` `THEREFORE m.=2m` `therefore`NEW magnification becomes DOUBLE. |
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| 34. |
In the question number 74, torque experienced by the system is |
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Answer» `10^(2)N` `therefore" "` Torque on dipole, `tau=PEsin180^(@)=0` |
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| 35. |
A thin film of a material of refractive index 1.38 is coated on a glass surface of refractive index 1.5.Light of wavelength 550 nm gives no reflected light. The minimum thickness of the film is : |
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Answer» `500 dotA` `2nt = ( m + 1/2) lambda` or `t = ((m + 1/2)lambda)/(2n)` ` = (lambda)/(4N)(for m = 0) = (5500)/(4 xx 1.38) Å` 
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| 36. |
You may be observed that, the fish inside the aquarium appears to be raised.What happens to the height of the object, (Thatvertically stands in the aquarium) when it isobserved by the fish |
| Answer» SOLUTION :Becomes taller. When light enters from rare to DENSER MEDIUM, it deviates TOWARDS the normal. | |
| 37. |
The bottom of water tank is lined with a plane mirror.A particle is projected from the origin O under gravity. The equation of the trajectory is y = a(x - x^2). The depth of the tank is 4a and this is completely filled. Find the distance of image with respect to mirror after reflection of rays from it. |
| Answer» SOLUTION :`(13A)/(3)` | |
| 38. |
Two identical capacitors are connected in series. Charge on each capacitor is q_0. A dielectric slab is now introduced between the plates of one of the capacitors so as to fill the gap, the battery remaining connected. The charge in each capacitor will now be |
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Answer» `(2q_0)/( 1+(1)/(K))` |
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| 39. |
A charged oil drop is suspended in unifrom of 3 xx 10^(4) V/m so that it neither falls nor rises. The charge on the drop will be ("take the mass of the charge " = 9.9 xx 10^(-15) kg & g = 10 m//s^(2)) |
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Answer» `3.3 XX 10^(-18)C` |
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| 40. |
Four identical mirror are made to stand vertically to form a square arrangement as shown in a top view. A ray starts from the midpoint M of mirror AD and after two reflections reaches corner D. Then, angle theta must be |
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Answer» `tan^(-1)(0.75)` From figure, `tan theta = (x)/((a//2)) "…."(i)` `tan theta = (a-x)/(y) "….."(ii) , tan theta = (a)/(a-y) "...."(iii)` From (i) and (ii) we get `(2X)/(a) = (a-x)/(y)` or `2xy = a^(2) - xa"......"(iv)` From (ii) and (iii), we get `(a-x)/(y) = (a)/(a-y) rArr 3xy = 2ay` (Using (iv)) `x = (2a)/(3)` Substituting this VALUE of x in equation (i), we get `tan theta = ((2a//3))/((a//2))= (4)/(3), :. cot theta = (1)/(tan theta) = 3/4` or `theta = cot^(-1)(0.75)` 
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| 41. |
If the potential difference between the anode and cathode of the X-ray tube is increases |
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Answer» The peaks at R and S would MOVE to SHORTER wavelength |
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| 42. |
Find the sum of magnitude of current (in Amp.) through R_(1), R_(2) and R_(3). All cells are ideal |
Answer» Solution : Potential of different points are shown. (i)current in `R_1` `I_1=(DELTAV)/R_1=(5-0)/1 A=5A` from left to right. (ii)current in `R_3` `I_3=(DeltaV)/R_3=30/1 A=30 A` from LOWER to HIGHER. (iii)For current in `R_2` using KCL `(10-x)/2+(0-x)/2+(0-x)/2+(-20-x)/1=0` `rArr 10/2-20=(3x)/2+x rArr x=-6V` `THEREFORE I_2=(20-6)/1A=14 A` |
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| 43. |
A large vessel is filled with a liquid. Two vertical plates touch the surlace of the liquid (Fig. 25.7). The dimensions of the plates are a and b, the distance between them is d. The plates have been charged by applying a voltage Po and then disconnected from the voltage source. To what height will the liquid rise? Ignore capillary effects. |
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Answer» Solution :When the liquid flows through a strongly non-uniform ELECTRIC field close to the edges of the plates, it is polarized and drawn into the SPACE between the plates. Since the charge on the plates remains unchanged in the process, and the capacitance of the capacitor increases, this is accompanied by a decrease in the energy of the field. This decrease is COMPENSATED by the increase in the potential energy of the column of líquid held between the plates. From the law of conservation of energy `(q^(2))/(2C_(0))=(q^(2))/(2C)+(mgh)/(2)` Here `C_(0)=(epsi_(0ab))/(d),C=(epsi_(0)b)/(d)[a+(epsi-1)h]` (see Problem 25.5) `m= rho bhd`. Substituting these values into the first formula we obtain, after some SIMPLIFICATIONS, `(epsi-1)q^(2)=epsi _(0)rho g a b^(2) [a+(epsi-1)h]` Express the charge on the plates in terms of the potential, `q=varphi_(0)C_(0)= epsi_(0)varphi_(0)ab//d` After simplifications we obtain a quadratic equation `h^(2)+(a)/(epsi-1)h-(epsi_(0)a varphi_(0)^(2))/(rho g d^(2))=0` Solving it we obtain the height the liquid rises. |
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| 44. |
रबी फसलें कौन- सी ऋतु में उगाई जाती है? |
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Answer» शीत ऋतु |
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| 45. |
Name and define the SI unit for the activity of a given sample of radioactive nuclei. |
| Answer» SOLUTION :See Point NUMBER 32 under the HEADING “CHAPTER At A Glance.” | |
| 46. |
A stone of mass m tied to a string of length l rotates along circumference of a circle with constant speed v. The torque on the stone is |
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Answer» zero |
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| 47. |
If two bulbs of 25 W and 100 W rated at 220 V are connected in series across 440 V supply. Will both the bulbs fuse? If not which one? |
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Answer» Solution :For an e electrical appliance `R=((V_(s)^(2))/(W))` `(R_(25))/(R_(100))=(W_(100))/(W_(25))=(100)/(25)=4"If"R_(100)-R, R_(25)=4R` `"in series,"V_(25)=(VxxR_(25))/(R_(25)+R_(100))440xx(4)/(5)=352V` `V_(100)=440-353=88V` From this it is clear that voltage across 100 W bulb is LESSER than specified (220 V). While across 25 W bulb is greater than (220V) so 25 W will FUSE. If the bulbs are connected in PARALLEL to 440 V supply, both bulbs will be fused because, applied voltage across each bulb is 440 V greater than specified voltage (220V) |
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| 48. |
A small satellite of mass m is revolving around earth in a circular orbit of radius r_(0) with speed v_(0) . At certain point of its orbit, the direction of motion of satellite is suddenly changed by angle theta = cos^(-1) (3/5) by turning its velocity vector , such that speed remains constant. The satellite consequently goes to elliptical orbit around earth. the ratio of speed at perigee to speed at apogee is |
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Answer» 3 |
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| 49. |
Two objects, one of mass 3 kg and moving with speed of 2 m/s and the other of mass 5 kg and speed 2 m/s, move toward each other and collide head-on. If the collision is perfectly inelastic, find the speed of the object after the collision. |
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Answer» 0.25m/s `m_(1)v_(1)+m_(2)v_(2)=(m_(1)+m_(2))v'` `v'=(m_(1)v_(1)+m_(2)v_(2))/(m_(1)+m_(2))` `=((3KG)(2m//s)+(5kg)(-2m//s))/(3kg+5kg)` `=-0.5m//s`. |
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