InterviewSolution
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.
| 1001. |
(4) 18100. The effective resistance between A and Bin the given circuit is3Ωq B30Ω(2) 7.52v(3), 4.7 Ω- |
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| 1002. |
Calculate the effective resistance between thepoints A and B in the circuit shown in Fig. 8.46. |
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Answer» now net resistance = 1+ 1/1 + 1 = 1+1+1= 3 ohm |
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| 1003. |
12) What do you mean by a polar molecule and a nonpolar molecule? Give one example of eacth. |
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Answer» A polar molecule has a netdipoleas a result of the opposing charges (i.e. having partial positive and partial negative charges) from polar bonds arranged asymmetrically.Water(H2O) is an example of a polar molecule since it has a slight positive charge on one side and a slight negative charge on the other. A molecule may be nonpolar either when there is an equal sharing of electrons between the two atoms of a diatomic molecule or because of the symmetrical arrangement of polar bonds in a more complex molecule. For example,boron trifluoride(BF3) has a trigonal planar arrangement of three polar bonds at 120°. |
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| 1004. |
(b) The mass of one molecule of a substance is 4.65x10. What is its molecular mass ? Whatcould the substance be? |
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Answer» thanks bro 😊😊😊 |
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| 1005. |
All the voluntary actions of our body are controlled byA. CerebrumB. MedulaC. PonsD. Cerebellum |
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| 1006. |
Why do farmers grow many fruits and vegetable crops inside largehouses? What are the advantages to the farmers? |
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Answer» As green house protects the plants to inside it from the climatic conditions outside and gives it suitable temperature for growth...that is why farmers do so.Through this they get many advantages....they have bountiful crops and their crops are also protected from birds and animals Farmersget many advantage like thecrops growwell..it also protect it from high blowing wind,and rodents.. Asgreen houseprotects the plants to inside it from the climatic conditions outside and gives it suitable temperature forgrowth...thatiswhyfarmers doso. A large greenhouse protect the plants to inside it from the climate conditions and gives it suitable temperature for the growth so that the farmer do this method. By the help of this method, they get many advantages are:1. They have good and pure crops.2. Their crops are protected from birds and animals. |
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| 1007. |
8.3.6 How can we separate a mixture of two miscible liquids? |
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Answer» Separating immiscible liquidsis done simply using aseparatingfunnel. Thetwo liquidsare put into the funnel and are left for a short time to settle out and formtwolayers. The tap of the funnel is opened and the bottomliquidis allowed to run. Thetwo liquidsare nowseparate. |
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| 1008. |
Give two examples of the substances whichexpand on heating. |
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Answer» grass trees plants etc 1. Any metal when heated expand.2. Plastic expand on heating as its particles get disassociate. |
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| 1009. |
Abodybodyofweightweight200sus200O N is suspended from a ceiling bys shówn in figure. Find the tensions T1 and31 Aoftwo ropes a2 in the two ropes45°2200N |
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| 1010. |
1. Name a few social evils prevailing in 19thcentury India. |
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Answer» The Indian society comprised of various social evils, malpractices and prejudices since the beginning of the ages. Some of them that were prevalent in the 19th century are discussed below: • System of giving dowry to the groom’s family by the bride’s family • Sati or immolation of the wife after the death of her husband in the same pyre with the husband • Child marriage • Female infanticide • Disapproval of re-marriage of windows and so on |
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| 1011. |
52.A mass 1 kg is suspended by a thread. It is() lifted up with an acceleration 4.9m/s2(i) lowered with an acceleration 4.9m/s2.The ratio of the tensions in two cases is |
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| 1012. |
4. A block of mass 02 kg is suspended from the ceiling bya light string. A second block of mass 0:3 kg is suspendedfrom the first block through another string. Find thetensions in the two strings. Take g-10 mvs |
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| 1013. |
12. Name any two substances that are responsible for causing biomagnificaiton.13.) Name the special tissue which nourish the Eb |
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Answer» ■DDT(dichlorodiphenyltrichloroethane) ■Hexachlorobenzene(HCB) ■PCBs(polychlorinated biphenyls) ■Toxaphene ■Monomethylmercury. |
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| 1014. |
32. A rectangular loop of sides 10 cm and 5 cmrying a current I of 12 A is placed in differententations as shown in the figures below:JEE Main 15 |
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Answer» option B is the correct answer using right hand curl rule.. upon curling the a the fingers the direction of thumb gives the direction of magnetic field. so, magnetic field is upward for B, which is correct. |
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| 1015. |
A block of mass 4 kg is kept on ground. The co-efficient of friction behween the block and the ground is0.80 An external force of magnitude 30 N is applied parallel to the ground The resultant force exertedby the ground on the block is: |
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Answer» To make the block move you need to apply force aboveùmgwhere u is 0.8So the minimum force comes out to be = 31.36 NewtonSo if u apply 30 Newton the ground will apply 30 NEWTON! when o say if u can do the question numerically then solve |
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| 1016. |
Cross-sectinal area (A)Cross-sectional area (A)Fig. 10.6 Cross-sectional areaFig. 10.7 Graph of R vs10.7 ResistivityFrom equations (10.1) and (10.2), we getRoc-(00.3)к is a constant which is called specific resistance of a conducting body or resistivity.Unit of resistivity.From eqn. (10.3),RA← L = 1 metre-Area of |
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| 1017. |
EXERCISES1 A steel wire of length 4.7 m and cross-sectional area 3.0 × 105 m23.5 m and cross-sectional area of 4.0same amount as a copper wire of lengthunder a given load. What is the ratio of the Young's modulus of steel to that o |
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| 1018. |
1. The two thigh bones (femurs), each of cross-sectional area 10 cm2, support the upper part ofa human body of mass 40 kg. What is the average pressure sustained by the femurs? |
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| 1019. |
n iron rod of 4 cm2 cross-sectional area isparallel to a magnetising field of 103 A/m. If themagnetic flux passing through it is 4x104 wb,then find permeability, relative permeability andmagnetic susceptibility of the substance. |
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Answer» First magnetic induction B = flux / area is to be found out. B = 4 x 10^-4/4 x 10-4= 1 TNow permeability μ = B / HGiven H = 1000 A/m, and found B = 1TSo μ = 1/1000 = 10-³H/m relative permeability is μ/μo = 10^-3/4π*10^-7 = 796.17Now with susceptibility chi we have to follow μ/μo = 1 + chiμo = 4 pi x 10-7 => chi = 10^-3/4π*10^-7 -1 = 796.17-1 = 795.17 |
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| 1020. |
9.1 A steel wire of length 4.7 m and cross-sectional area 3.0 x 10-5 m2 stretches by thesame amount as a copper wire of length 3.5 m and cross-sectional area of 4.0 x 10-5 m2under a given load. What is the ratio of the Young's modulus of steel to that of copper? |
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| 1021. |
Tuo Ο DIFFERENT (iguiels.HAv1.N(nSAME MASS ANE MxENTOEATHER IRA RE RESPECTIVELYー0.gsx \03THE DENSITY OF THEIR MİXTORE |
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| 1022. |
31. Consider the situation shown in figure (8-E2). The systemis released from rest and the block of mass 10 kg isfound to have a speed 03 m/s after it has descendedthrough a distance of 1 m. Find the coefficient of kineticfriction between the block and the table.4.0kg1.0kg□ |
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| 1023. |
mational highway longest |
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Answer» NH 44covers the North-South Corridor of NHDP and it is officially listed as running over 3,745 km (2,327 mi) from Srinagar toKanyakumari. It is the longest national highway in India |
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| 1024. |
Cross section area of a steel wire (Y=2.0 × 10^11 N/m^2is 0.1 cm^2. The required force, to make its lengthdouble will be |
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| 1025. |
A steel wire of length 2.5 m and area ofcross section 1.25 mm2 when it isstretched by a force of 40 N and Y fosteel is 2x1011 N/m. The extensionproduced in a steel wire is |
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| 1026. |
he Young's modulus of brass and steel are respectively 10m2 A brass wire and a100 N/m2 and 2x 10-osteel wire of the same length are extended by 1 mm under thene force, the radii of brass and steel wires arc Ri: and Rs respectively. Thenb) RsaN24 |
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| 1027. |
bending towards light |
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Answer» Thisbending toward lightis called phototropism. |
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| 1028. |
We can see non-luminous object when lighta. Light emitted by the object falls on the eye.b. Is reflected from the object towards our eyec. Completely passes through the objectd. Gets completely absorbed by the object.3. |
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Answer» In order for us to see anobjectthat isnon-luminous, it must reflect some of the light it receives from aluminoussource, such as the Sun. Most of theobjectswe see, such as cars, clouds or even the Moon, are notluminous; it is just that they reflect sunlight. |
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| 1029. |
8. A point charge 4. = 400uC exerts a forceof (601 - 80j)N on an other charge qz invacuum separated by a distance of 30m. Findthe charge 42. (2.5 x 10-20) |
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| 1030. |
The given magnet is divided into three partsand CName the parts where the strength of themagnetic field is |
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Answer» Maximum magnetic field is present in part A and C Minimum magnetic field is present in region B this is because the part A and C are magnetic poles so that these parts have maximum number of magnetic field lines which determine the intensity of magnetic field and B is the center of the magnet which has no magnetic field lines. Therefore intensity of magnetic field near B is almost 0 |
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| 1031. |
b) The given magnet is divided into three parts A, B, and C.AİBİCName the parts where the strength of the magnetic field is:(i) maximum(ii) minimum |
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Answer» i) Maximum magnetic field is in the region A and C. (ii) Minimum magnetic field is in the region B. This is because A and C are magnetic poles and have maximum number of magnetic field lines which determine the intensity of magnetic field while B is centre of the magnet that has no magnetic field lines. So intensity of magnetic field near B is almost zero. please like the solution 👍 ✔️👍 |
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| 1032. |
ohm-metre.Art. 7 APPLICATIONS OF ELECTRICITY |
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| 1033. |
If W is work done on a liquid film having area A, then Surface energy is (a) Wx L, (b) WxA, (c)W/L(d) W/A8. |
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Answer» surface energy is = work done per unit area = W/A |
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| 1034. |
1.14 LARGE-SIGNAL OPERATION OE OP-AMPs |
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Answer» please soecify the question clearly An operational amplifier is a DC-coupled high-gain electronic voltage amplifier with a differential input and, usually, a single-ended output. |
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| 1035. |
4) All of theseWhen a force of constant magnitude and a fixeddirection acts on a moving object, then its path is(1) Circular(2) Parabolic(3) Straight line(4) Either (2) or (3)akash Educational Services Pvt. Ltd.-Regd. Office: Aakash T |
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Answer» option 4 is correct if the force acts parallel to the direction of movement , the acceleration provided will be in the same direction..so it will follow straight line and if the force is not parallel.. then it will be combined movement of curve+ linear, so parabola |
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| 1036. |
6. किसी " लंबाई तथा 'A' मोटाई के तार का प्रतिरोध 82 है। इसी पदार्थ के किसी अन्य तार का प्रतिरोध क्या होगा ।जिसकी लंबाई है तथा मोटाई 2A है ?A Wire of given material having length T and area of cross-section has aresistance of 89. What would be the resistance of another wire of the same materialhaving length and area of cross-section 2A. |
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Answer» We know thatR= pl/A= 8nowR is proportional to l/AnowR= p(l/2)/2A = R= p(0.5l)/2A= 8*0.5/2= 40/20= 2 ohm |
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| 1037. |
A conductor having diameter 0.4 mmhas resistivity 5 * 10^-7 ohm metreand it's length is 10.05 metre. What isthe resistance of the conductor? |
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| 1038. |
3. Calculate the resistance of a copper wire 1.0 Km long and 0.50 mm diameter if theresistivity of copper is 1.7 x 10-8 2m |
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| 1039. |
Surface tension of liquid is 5N/m. If a thin flm of area 0.02 m2 is made in a ring then the surface energy wllbe(A) 16x 103 J(B) 2.5 10-3 J(C) 1 x 10-1 J(D) 5 x 10-J |
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Answer» W=T×ΔA=5×2×(0.02) (Film has two free surfaces) =2×10−1J The answer to option C is 2×10^-1 J . The option provided is wrong. |
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| 1040. |
A Wire of given material having length 7' and area of cross-section 'A' has aresistance of 8Ω. What would be the resistance of another wire of the same materialhaving length and area of cross- section 2A.2 |
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Answer» Resistance = p l/a [ p is constant]For first wire length = l and area of croos section = aResistance = p l/a = 8 ohm [given] For second wire length = l/2 and area of cross section = 2aResistance = p (l/2) / 2a = p l/4a = p l/a * 1/4 [p l/a = 8 ohm] = 8*1/4 = 2 ohm The resistance of second wire = 2 ohm |
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| 1041. |
2A Wire of. given material having length T' and area of cross-section A' has aresistance of 82. What would be the resistance of another wire of the same materialhaving lengthand area of cross- section 2.A.2 |
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Answer» We know that the resistance is P(rho) l/a. We can ignore P over here because the material is same in both cases. Now, R=l/a. 8=l/a. In the second case, they are asking resistance if the area is double and length is halved. So, R=l/2 whole divided by 2a.We get R=l/4a. But we l/a=4. So, 4/4=1 and so the answer is 1 ohm. |
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| 1042. |
5.Calculate the resistance of 50m length of wire of cross section area 0.01 millimetresquare and of resistivity 5x108 ohm meter? |
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Answer» resistance is R = ρl/a so , here ρ = 5*10^-8 and l = 50 , and a = 0.01mm² = 0.01*10⁻⁶ m² so, R = (5*10^-8)*50/0.01*10^⁻⁶R = 250 Ω |
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| 1043. |
The resistance of a wire of length 300 m and cross-section area 1.0 mm made of material of resistivity10 x 10 Sam is: |
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| 1044. |
The normal systolic blood pressure isA. 80 mmB. 100 cmC. 120 mmD. 130mm |
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Answer» C. 120 mm Explaination : systolic blood pressure, measures the pressure in our blood vessels when heart beats. |
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| 1045. |
4.A. 25C. 25The least distance of distinct vision for the normal vision isB.25A. 25mB. 25cmC. 25mmD. of these |
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Answer» B. 25 cm Explaination : Theleastdistance ofvisionfor anormalyoung eye is 25 cm. Theleastdistance of distinct visionis theminimumdistance of an object to see clear and distinct image. |
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| 1046. |
Additional Exercises5.24 Figure 5.17 shows the position-time graph of a body of mass 0.04 kg. Suggest asuitable physical context for this motion. What is the time between two consecutiveimpulses received by the body? What is the magnitude of each impulse ?x (cm)2 4 6 8 10 12 14 16 t(s)Fig. 5.17 |
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| 1047. |
The potential energy of a certain spring when stretchedthrough a distance z is 10 J. What is the amount of workdone on the same spring to stretch it through anadditional distance x? [AFMC 1991] IAns. 30 J24 |
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| 1048. |
8. The work done for a certain spring when stretchedthrough a distance S is 10 joule. The amount of work(in joule) that must be done on this spring to stretch itthrough an additional distance S will be(a) 30(b) 40(c) 10(d) 20. |
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| 1049. |
iesWrire two postulates of Bohr model of an atom based on quantun cne |
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Answer» The main and most important postulates of Bohr’s theory : Postulates of Bohr's theory of hydrogen atom 1. NUCLEAR CONCEPT :An atom consists of a small and massive central core, called nucleus around which planetary electrons revolve. The centripetal force required for their rotation is provided by the electrostatic attraction between the electrons and the nucleus. 2. QUANTUM CONDITION :Of all the possible circular orbits allowed by the classical theory, the electrons are permitted to circulate only in those orbits in which the angular momentum of an election is an integral multiple of h/2 pye, h being Planks constant. Therefore, for any permitted orbit, L=mvr =nh/2PYE, n=1,2,3…… WheWhere L, m, and v are the angular momentum, mass and speed of the electron, r is the radius of the permitted orbit and n is positive integer called principal quantum no. The above equation is Bohr’s famous quantum condition. |
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| 1050. |
7. In the Bohr model of hydrogen atom, the electronmakes about 0.6 x 10^16 rev /s around the nucleus.What is the average current at a point on the orbit ofelectron ? e=1.59x 10-19 C |
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