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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.
| 901. |
A massless, inextensible string of length 1 m has a breaking strength of 1 kg wt. A stone of mass 0.2 kg tied to one end of the string is made to move in a vertical circel. Can te stone decribe the vertical circle? `(g=10 ms^(-2))` |
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Answer» `T_(max)=1 kg.wt=1xx10=10N` Let tension at towest point be T `T_(min)=6mg` `T_(min)=6mg=6xx0.2xx10=12NgtT_(min)` So, the stone cannot describe the vertical circel. |
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| 902. |
Two identical `5 kg` blocks are moving with same speed of `2 m//s` towards each other along a friction less horizontal surface . The two blocks collide , stick together and come to rest. Consider the two blocks as a system . The work done by the external and ijnternal force are respectively: |
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Answer» Here, `m_(1)=m_(2)=5kg,` `v_(1)=2m//s, v_(2)=-2m//s` Final velocity `v=0` As no external forces are involves, therefore, `vec(F)_(ext)=0, W_(ext)=vec(F)_(ext).vec(s)=Zero` According to work energy theorem, Total work done `=` Change is KE `W_("ext")+W_("int")=` final KE - inital KE `=0-((1)/(2)m_(1)v_(1)^(2)+(1)/(2)m_(2)v_(2)^(2))` `0+W_(i nt)=-(1)/(2)[5(2)^(2)+5(-2)^(2)]=-20J` `:. W_("int")=-20J` Negative sign show that internal forces of action and reaction act on the two blocks in a direction opposite to their motion. |
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| 903. |
A small block of mass 100 g is pressed again a horizontal spring fixed at one end to compress the sprign through 5.0 cm. The spring constant is 100 N/m. When released, the block moves hroizontally till it leaves the spring. Where will it hit the ground 2 m below the spring? |
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Answer» When the block is released, it moves horizontally with speed V till it leaves the spring. By energy conservation, `1/2kx^2=1/2mV^2` `V^2=(kx^2)/(m)impliesV=sqrt((kx^2)/(m))` Time of flight, `t=sqrt((2H)/(g))` So horizontal distance travelled from the free end of the spring is `Vxxt=sqrt((kx^2)/(m))xxsqrt((2H)/(g)a)` `=sqrt((100xx(0.05)^2)/(0.1))xxsqrt((2xx2)/(10))=1m` So at a horizontal distance of `1m` from the free end of the spring. |
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| 904. |
An ice cube of size `a = 10 cm` is floating in a tank (base area `A = 50 cm xx 50 cm`) partially filled with water. The change in gravitational potential energy, when ice melts completely is (density of ice is `900 kg//m^(2)`)A. `-0.072` JB. `-0.24` JC. `-0.016` JD. `-0.045` J |
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Answer» Correct Answer - D |
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| 905. |
Three components of a force acting on a particle are varying according to the graphs as shown. To reach at point B(8, 20, 0)m from point A(0, 5, 12)m the particle moves on paths parallel to x-axis then y-axis and then z-axis, then work done by this force is: A. 192 JB. 58 JC. 250 JD. 125 J |
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Answer» Correct Answer - C |
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| 906. |
A body of mass `m kg` lifted by a man to a height of one metre in `30 sec` . Another mass lifted the same mass to the same height in `60 sec` . The work done by then are them are in the ratio.A. `1:2`B. `1:1`C. `2:1`D. `4:1` |
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Answer» Correct Answer - B |
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| 907. |
Examine table.s and express a) The energy required to break one bond in DNA in eV, b) the kinetic energy of an air molecule `(10^(-2)` J in eV, c) The daily intake of a human adult in kilocalories. |
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Answer» a) Energy required to break one bond of DNA is: `(10^(-20)/(1.6 xx 10^(-19) J//eV) =~ 0.06 eV` Note, 0.1eV = 100meV (100 millielectron volt). B) the kinetic energy of an air molecule is `(10^(-21))/(1.6 xx 10^(-19)J//.eV=~ 0.0062eV` this is the same as 6.2 meV. c) The average human consumption in a day is `10^(7)/(4.2xx 10^(3)J/kcal) -~2400 kcal` We point out a common misconception created by newspapers and magzines. They mention food values in calories and urge us to restrict diet intake to below 2400 claories. What they should be saying in kilocalories (kcal) and not calories. A person consuming 2400 calories a day will soon starve to death 1 food calorie is 1 kcal. |
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| 908. |
Express the enenrgy of big bang in eV |
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Answer» Energy of big bang `=10^(68)J` `1.6xx10^(-19)J=1 eV` Energy of big bang `=(10^(68)J)/(1.6xx10^(-19)J//eV)=6.25xx10^(68)eV` |
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| 909. |
Express the daily intake of a human adult in kiloc alorie |
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Answer» The average human consumption in a day `=10^(7)J` `1. cal =4.2J` `1 K cal =4.2xx10^(3)J` The average human consumption in a day `=(10^(7)J)/(4.2xx10^(3)J//Kcal)` `=2400 kcal` |
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| 910. |
A body is mass `300 kg` is moved through `10 m` along a smooth inclined plane of inclination angle `30^@`. The work done in moving (in joules) is `(g = 9.8 ms^(-2))`.A. 4900B. 9800C. 14700D. 2450 |
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Answer» Correct Answer - D (d) We know that, `W=mg cos theta xxs` `W=300xx9.8xxcos30^(@)xx10` `=300xx9.8xx(sqrt3)/(2)xx10=2450J` |
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| 911. |
A rigid body moves a distance of 10m along a straight line under the action of a force 5N. If the work done by this force on the body is 25 joules, the angle which the force makes with the force makes with the direction of motion of the body is: |
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Answer» Work is measured by the product of the applied force and the displacement of the body in the direction of the force. `:. "Work" = "Applied force xx Displacement"` `W=(F cos theta)xx s=Fscos theta` Given, W=25 J,F=5 N,s=10m `:. " "cos theta=(W)/(F.s)=(25)/(5xx10)=(1)/(2)` `implies " "theta =cos^(-1)((1)/(2))=60^(@)` Hence,angle between force and direction of motion of the body is `60^(@)`. |
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| 912. |
A particle moves from point P(1,2,3) to (2,1,4) under the action of a constant force `F=(2hati+hatj+hatk)N`. Work done by the force isA. 2 JB. 4 JC. 16 JD. 8 J |
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Answer» Correct Answer - A (a) Particle moves from P(1,2,3) to (2,1,4) `:. "Distance" PQ =(2hati+hatj+4hatk)-(hati+2hat+3hatk)` `d=PQ=(hati-hatj+hatk)` Force `F=(2hati+hatj+hatk)` Work done by the force `W=F.d=(2hati+hatj+hatk).(hati-hatj+hatk)=2-1+1=2 J` |
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| 913. |
For Q. 25 find the minimum constant force to be applied to body of mass `m_(1)` in order to shift body of mass `m_(2).` Take `m_(1)=m_(2)=10kg, mu=0.5` |
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Answer» `F.x=mum_(1)gx-1/2kx^(2)=0" "...(i)` But `Kx=mu_(2)g.` (for just moving body of mass `m_(2))` Put the value of K in (i) to get F. F.x `-mum_(1)gx-1/2kx^(2)=0` But `kx=mum_(2)g` (for moving body of mass `m_(2)`) `Fx-mum_(1)gx-1/2((mum_(2)g)/(x))x^(2)=0` `F=mum_(1)g+1/2mum_(2)g` `=mu(m_(1)+(m_(2))/(2))g` `=0.5(10+10/2)10` `=75N` |
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| 914. |
Which of the following statements is correctA. Work done by static friction is always zeroB. Work done by kinetic friction is always negativeC. The negative of t he work done by the conservative internal fores on a system equals the change in kinetic energyD. The work dony by all the forces on a system equals the change in kinetic energy |
| Answer» Correct Answer - D | |
| 915. |
A block of mass m is taken from A to B under the action of a constant force F. Work done by this force is A. `(piFR)/(2)`B. `piFR`C. FRD. Zero |
| Answer» Correct Answer - C | |
| 916. |
Two bodies of masses `m_(1)and m_(2)` are connected by a non-deformed light speing and lie on a horizontal plane. Coefficient of friction between the surface and the blocks is `mu.` Find an expression for the work done by the various foeces. Force F is applied horzontally to body of mass `m_(1)` in order to shift body of mass `m_(2).` |
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Answer» Let the motion of `m_(2)` start when the spring is strefched by x slowlt. Work done by various forces will involve 3 terms, F.x `(-um_(1)g)x and -1/2kx^(2)` Let the motion of `m_(2)` start when one spring is solwe stretched by x slowly Work done by various forces `=F.x-(mum_(1)g)x-1/2kx^(2)` =Change in KE=0 Where k is force constant of the spring |
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| 917. |
Two particles, each of mass m are attached to the two ends of a light string of length l which passes through a hole at the centre of a table. One particle describes a circle on the table with angular velocity `omega_(1)` and the other describes a circle as a conical pendulum with angular velocity `omega_(2)` below the table as shown in figure -3.86. if `l_(1)` and `l_(2)` are the lengths of the portion of the string above and below the table then : A. `l_(1)/l_(2)=(omega_(2)^(2))/(omega_(1)^(2))`B. `l_(1)/l_(2)=(omega_(1)^(2))/(omega_(2)^(2))`C. `1/(omega_(1)^(2))+1/(omega_(2)^(2)=l/g`D. `1/(omega_(1)^(2))+1/(omega_(2)^(2)) gt l/g` |
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Answer» Correct Answer - A |
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| 918. |
How much work must work be done by a force on 50 kg body in order to accelerate it in the direction of force from rest to `20 ms^(-1)` is 10 s?A. `10^(-3)`JB. `10^(4) J`C. `2xx10^(3) J`D. `4xx10^(4) J` |
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Answer» Correct Answer - B (b) As, equation of motion is `v=u+at implies20=0+axx10` `20=axx10 implies a=2 m//s^(2)` Now, distance `s=ut+(1)/(2)at^(2)` `implies s=0+(1)/(2)xx2xx10xx10` or `" " s=100 m W=F.s = mas = 50xx2xx100=10^(4) J` |
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| 919. |
Assertion: The instantcous power of an agent is measured as the dot product of instaneous velocity and the force acting on it at that instant. Reason: The unit of instaneous power is watt.A. If both assertion and reason are true and reason is a true explanation of assertion.B. If both assertion and reason are true but reason is not the correct explanation of assertion.C. If assertion is true but reason is false.D. If both assertion is and reason are false. |
| Answer» `P = bar F .bar v and unit of power is wan | |
| 920. |
Given that a force F acts on a body for time `t_(1)` and displaces the body by `vec d` . In which of the following cases the velocity of the body must increases ?A. `F gt d`B. `F lt d`C. F||d`D. hatF _|_ hatd ` |
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Answer» Correct Answer - C In this case work will be done by the force which will increase the kinetic energy. |
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| 921. |
Assertion: According to law of conservation of machainical energy change in potential energy is equal and opposite to the change in kinetic energy Reason: Mechanical energy is not a conserved quantity.A. If both assertion and reason are true and the reason is the correct explanation of the assertionB. If both assertion and reason are true but reason is not the correct explanation of the assertionC. If asserti on is true but reason is falseD. If the assertion and reason both are false |
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Answer» Correct Answer - C |
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| 922. |
Assertion: According to law of conservation of machainical energy change in potential energy is equal and opposite to the change in kinetic energy Reason: Mechanical energy is not a conserved quantity.A. If both assertion and reason are true and reason is a true explanation of assertion.B. If both assertion and reason are true but reason is not the correct explanation of assertion.C. If assertion is true but reason is false.D. If both assertion is and reason are false. |
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Answer» For conservative force the state of kinetic and potential energies at point remain constant throughout the motion This is knoen as law of coneservitive of metanical energy According to this law, Kinetic energy + potential energy = constant or `Delta E+ Delta U = 0 or Delta K+ Delta U` |
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| 923. |
Assertion: According to law of conservation of machainical energy change in potential energy is equal and opposite to the change in kinetic energy Reason: Mechanical energy is not a conserved quantity.A. If both assertion and reason are true and reason is a the correct explanation of assertion.B. If both assertion and reason are true but reason is not the correct explanation of assertion.C. If assertion is true but reason is false.D. If assertion is true and reason are false. |
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Answer» For conservative force the sum of kinetic and potential energies at any point remain constant throughout the motion This is knoen as law of coneservitive of metanical energy According to this law, Kinetic energy + potential energy = constant or `Delta K + Delta U = 0 or Delta K = - Delta U` |
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| 924. |
Assertion: According to law of conservation of machainical energy change in potential energy is equal and opposite to the change in kinetic energy Reason: Mechanical energy is not a conserved quantity.A. If both, Assertion and Reason are true and Reason is correct explantion of Assertion.B. If both, Assertion and Reason are true but Reason is not a correct explanation of the Assertion.C. If Assertion is true but the Reason is false.D. If both, Assertion and Reason are false. |
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Answer» Correct Answer - C Here, Assertion is true, but the reason is false. Accroding to law of conservation fo mechanical energy `(` If no external forces do work on a system `)`, for internal conservation forces, `K+U=`constant or ``DeltaK+DeltaU=0` or `DeltaK=-DeltaU` |
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| 925. |
An automobile of mass m accelerates, starting from rest. The engine supplies constant power P, show that the velocity is given as a function of time by `v=((2Pt)/(m))^(1//2)` |
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Answer» Power,P=constant Work done up to time t is W =Pt From work energy theorem, `W=DeltaKE " or "Pt=(1)/(2)mv^(2)` `:. " " v=((2Pt)/(m))^(1//2)` |
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| 926. |
A pump can take out 7200 kg of water per hour from a well 100 m. deep. The power of pump, assuming its efficiency as `50%` will be |
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Answer» `"Output power"=(mgh)/(t)=(7200xx10xx100)/(3600)=2000 W` `"Efficiency" eta =("output power")/("input power")` `"Input power"=("output power")/(eta)=(2000xx100)/(50)=4 kW` |
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| 927. |
A spring gun has a spring constant of `80Ncm^(-1)`. The spring is compressed `12cm` by a ball of mass `15g`. How much is the potential energy of the spring ? If the trigger is pulled, what will be the velocity of ball ? |
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Answer» Here, `K=80Ncm^(-1)=80xx100Nm^(-1)` `x=12cm=12xx10^(-2)m,` `m=15g=15xx10^(-3)kg` `P.E` of spring `=(1)/(2)Kx^(2)` `=(1)/(2)xx80xx100(12xx10^(-2))^(2)=57.6J` Again, KE of ball `= P.E.` of spring `(1)/(2)m upsilon^(2)=57.6, upsilon=sqrt((2xx57.6)/(15xx10^(-3)))=sqrt(7680)` `-87.6m//s` |
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| 928. |
Identify the correct statement about work energy theorem.A. Work done by all the forces is equal to the decrease in potential energy.B. Work done by all the forces except the conservative is equal to the change in mechanical energy.C. Work done by all the forces is equal to the change in kinetic energy .D. Work done by all the forces is equal to the change in potential energy |
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Answer» Correct Answer - B::C Work done by conservative force `=-DeltaU`, `Work done by all the forces `=Delta K` Work done by forces other than conservative forces `=Delta E`. |
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| 929. |
The PE of a certain spring when streched from natural length through a distance `0.3m` is `5.6J`. Find the amount of work in joule that must be done on this spring to stretch it through an additional distance `0.15m`. |
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Answer» Correct Answer - `(7)` `5.6=1/2k(0.3)^2` `W=1/2k[(0.3+0.15)^2-0.3^2]` Solving to get: `W=7J` |
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| 930. |
Underline the correct alternative:Work done by a body against friction always results in a loss of its kinetic/potential energy. |
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Answer» Kinetic energy The work done against the direction of friction reduces the velocity of a body. Hence, there is a loss of kinetic energy of the body. |
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| 931. |
A uniform force of (\(2\hat i + \hat j\)) + N acts on a particle of mass 1 kg. The particle displaces from position(\(3 \hat j +\hat k\)) m to (\(5\hat i + 3\hat j\)) m. Th e work done by the force on the particle is(a) 9 J (b) 6 J (c) 10 J (d) 12 J |
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Answer» Correct answer is (c) 10 J Given, Force, F = \(2 \hat i + \hat j\) \(\vec r_1\) = (\(3 \hat j + \hat k\)) m \(\vec r_2\) = (\(5 \hat i + 3\hat j\)) m Therefore, \(\vec S\) = \(\vec r_2\) - \(\vec r_1\) = (\(5 \hat i + 3\hat j\)) - (\(3 \hat j + \hat k\)) = \(5 \hat i + 3\hat j\) - \(3 \hat j - \hat k\) = ( \(5 \hat i - \hat k\)) m Therefore, \(W = \vec F.\vec S\) = (\(2 \hat i + \hat j\)).( \(5 \hat i - \hat k\)) = 10 J |
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| 932. |
This question has statement `1` and statement `2` . Of the four choice given after the Statement , choose the one that best describe the two Statement . If the spring `S_(1) `and `S_(2)` of force constant `k_(1)` and `k_(2)` respectively , are streached by the same force , it is found that more work is done on spring `S_(1)` then on spring `S_(2)` Statement -1: If statement by the same answer work done on `S_(1)` work on `S_(1)` is more then `S_(2)` Statement - 2 : `k_(1) ltk_(2)`A. Statement - 1 is ture , statement -2 is true, and statement-2 is correct explanation of statement -1.B. Statement-1 if ture, statement-2 is true, but statement -2 is not a correct explanation of statement-1.C. Statement-1 is true, but statement -2 is false.D. Statement-1 is false, but statement-2 is true. |
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Answer» Correct Answer - D Let `x_(1)` be extension produced in `S_(1)` and `x_(2)` be extension produced in `S_(2)`, when two springs are stretched by the same force. `:. F=K_(1)x_(1)=K_(2)x_(2)` `W_(1)=Fx_(1)` and `W_(2)=Fx_(2)` As`W_(1) gt W_(2),Fx_(1) gt Fx_(2)` or `x_(1) gt x_(2)` `:. K_(1) lt K_(2)` Statement 2 is true. When two springs are stretched by the same amount, `i.e., x_(1)=x_(2)=x` `W_(1)=(1)/(2) K_(1)x^(2)` As `K_(1) lt K_(2), W_(1) lt W_(2)`. Statement 1 is false. |
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| 933. |
Statement-1 : A body with negative energy cannot have momentum. Statement-2 : This is because momentum can be positive only.A. Statement - 1 is ture , statement -2 is true, and statement-2 is correct explanation of statement -1.B. Statement-1 if ture, statement-2 is true, but statement -2 is not a correct explanation of statement-1.C. Statement-1 is false, but statement-2 is false.D. Statement-1 is false, but statement-2 is true. |
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Answer» Correct Answer - C Statement`-1` is false. Negative energy means, `K+U=` negative `:. K!=0.` Therefore, the body can have linear momentum. Statement `-2` is false. Choice `(c)` is correct. |
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| 934. |
A force `F=2x+3x^2`N acts on a particle in the x-direction. Find the work done by this force during a displacement `x=1.0m ` to `x=2.0m` |
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Answer» Here force is variable `W=int_(x1)^(x2)Fdx=int_1^2(2x+3x^2)dx` `=2.(x^2)/(2)+3(x^3)/(3)=|x^2+x^3|_1^2` `={(2)^2+(2)^3}-{1^2+1^3}=12-2=10J` |
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| 935. |
The first ball of mass `m` moving with the velocity `upsilon` collides head on with the second ball of mass `m` at rest. If the coefficient of restitution is `e`, then the ratio of the velocities of the first and the second ball after the collision isA. `(1-e)/(1+e)`B. `(1+e)/(1-e)`C. `(1+e)/(2)`D. `(1-e)/(2)` |
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Answer» Correct Answer - A Here, `m_(1)=m_(2)=m, u_(1)=u, u_(2)=0`. Let `upsilon_(1), upsilon_(2)` be their velocities after collision. According to principle of conservation of linear momentum. `m u +0=m(upsilon_(1)+upsilon_(2))` or `upsilon_(1)+upsilon_(2)=u ….(1)` By definition, `e=(upsilon_(2)-upsilon_(1))/(u-0)` or `upsilon_(2)-upsilon_(1)= e u .....(ii)` Add `(i)` and `(ii)``upsilon_(2)=(u(1+e))/(2)` Subtract `(ii)` from `(i)` `upsilon_(1)=((1-e)u)/(2) :. (upsilon_(1))/(upsilon_(2))=(1-e)/(1+e)` |
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| 936. |
The first ball of mass `m` moving with the velocity `upsilon` collides head on with the second ball of mass `m` at rest. If the coefficient of restitution is `e`, then the ratio of the velocities of the first and the second ball after the collision is |
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Answer» Here, `m_(1)=m_(2)=m,u_(1)=u, u_(2)=0` Let `upsilon_(1)` and `upsilon_(2)` be their respective velocities after collision. As `m_(1)u_(2)+m_(2)u_(2)=m_(1)upsilon_(1)+m_(2)upsilon_(2)` `:. m u+0=m(upsilon_(1)+upsilon_(2)) ` or `u=upsilon_(1)+upsilon_(2)` ...(i) Again, by definition, `e=(upsilon_(2)-upsilon_(1))/(u-0)` `:. upsilon_(2)-upsilon_(1)=ue` Solving (i) and (ii), we get `upsilon_(2)=((1+e)u)/(2)` `upsilon_(1)=((1-e)u)/(2) :. (upsilon_(1))/(upsilon_(2))=(1=e)/(1+e)` |
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| 937. |
A force act on a `30 gm` particle as a friction of the particle as a function as given by `x = 3 t - 4 t^(2) + t^(3)`, where `x` is in metros and `t` is in seconds. The work done during the first `4` second isA. 5.28 JB. 450 mJC. 490 mJD. 530 mJ |
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Answer» Correct Answer - A `x=3t-4t^2+t^3` `v=(dx)/(dt)=3-8t+3t^2` `t=0`, `v_1=3(m)/(s)` `t=4s`,`v_2=3-8(4)+3(4)^2` `=3-32+48=19(m)/(s)` `W_(1rarr2)=K_2-K_1=(1)/(2)m(v_2^2-v_1^2)` `=(1)/(2)xx(30)/(1000)(19^2-3^2)` `=0.015xx22xx16=5.28J` |
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| 938. |
In generation of hydroelectric power, water falls from some height and gravitational potential energy of water is converted into electric power with the help of turbine. In one such event, `7.2xx10^4kg` of water falls per hour from height 100m and half of the gravitational potential energy is converted into electric energy. How much power is generated? |
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Answer» Lost potential energy`=mgh` `=7.2 xx10^(4)xx10xx100` `=7.2xx10^(7)J` Electric energy produced `=(1)/(2)xx7.2xx10^(7)=3.6xx10^(7)` Power produced `P=(3.6xx10^(7))/(t)=(3.6xx10^(7))/(3600)=10^(4)W` |
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| 939. |
A block is placed on the top of a plane inclined at `37^@` with horizontal. The length of the plane is `5m`. The block slides down the plane and reaches the bottom. a. Find the speed of the block at the bottom if the inclined plane is smooth. b. Find the speed of the block at the bottom if the coefficient of friction is `0.25`. |
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Answer» Let h be the height of the inclined plane. `h=5sin37^@=3m` ltbgt a. As the block slides down the inclined plane, it loses gravitational potential energy and gains KE. Loss in GPE=gain in KE mg (loss in height) `=KE_j-KE_i` `implies mgh=1/2mv^2-0` `impliesv=sqrt(2gh)=sqrt(2xx9.8xx3)=7.67ms^-1` b. As the blocks comes down, it loses GPE. It gains KE and does work against friction. Loss in GPE=gain in KE+work done against friction `=mgh=(1/2mv^2-0)+(mumgcos37^@)s` `=3mg=1/2mv^2+(0.25)xxmgxx4/5xx5` `impliesv=sqrt(4g)=6.26ms^-1` |
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| 940. |
A conservative force held function is given by `F=k//r^2`, where k is a constant. a. Determine the potential energy funciton `U(r)` assuming zero potential energy at `r=r_0`. b. Also, determine the potential energy at `r=oo`. |
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Answer» a. Using the definition of potential energy function, `U(r)-U(r_0)=-int_(r_0)^rFdr` `U(r)-U(r_0)=-kint_(r_0)^r(dr)/(r^2)=k[1/r]_(r_0)^r=k[1/r-1/r_0]` Since at `r=r_0`, `U(r_0)=0`, therefore, `U(r)=k/r-k/r_0` b. Potential energy at `r=oo` is `U_(oo)=-k//r_0` |
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| 941. |
Explain the social changes in London which led to the need for the Underground railway. Why was the development of the Underground criticised? |
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Answer» The development of suburbs as a part of the drive to decongest London led to the extension of the city beyond the range where people could walk to work. Though these suburbs had been built, the people could not be persuaded to leave the city and stay far away from their places of work in the absence of some form of public transport. The Underground railway was constructed to solve this housing problem. |
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| 942. |
Explain what is meant by the Haussmanisation of Paris. To what extent would you support or oppose this form of development? Write a letter to the editor of a newspaper, to either support or oppose this, giving reasons for your view. |
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Answer» Haussmanisation of Paris refers to the re-building of Paris by Baron Haussmann in the mid-eighteenth century. When Louis Napoleon III came to power, he appointed Haussmann as the chief architect of the new city. He laid out new streets, straight sidewalks, boulevards and open avenues, and planted full-grown trees. Haussmann’s architectural plans had positives as well as negatives. His name has become a representation of forcible reconstruction to enhance the beauty of a city and impose order. This is because his plans led to the displacement of 350,000 people from the centre of Paris. This included many poor people who were now rendered homeless. |
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| 943. |
A force F is acted on a body as shown in the figure. Find the impulse of the force in first four seconds. |
| Answer» Impulse can be calculated as area under force time graph. Therefore, impulse =12 Ns | |
| 944. |
A uniform speing is kept on a horizontal frictionless floor such that one end is fixed to a vertical wall and other end if free. Mass of the speing is m. At the instant. Shown in figure veloity of free end is v and length of the speing is L. Assumign that speed varies linearly from zero to v, the kinetic energy of the spring will be A. `1/2mv^(2)`B. `1/4mv^(2)`C. `1/6mv^(2)`D. `1/8mv^(2)` |
| Answer» Correct Answer - C | |
| 945. |
A particles of mass m is fixed to one end of a light spring of force constant k and unstreatched length l. the system is rotated about the other end of the spring with an angular velocity `omega` in gravity free space. The increase in length of the spring is A. `(momega^(2)l)/k`B. `(momega^(2)l)/(k-m omega^(2))`C. `(momega^(2)l)/(k+m omega^(2))`D. None of these |
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Answer» Correct Answer - B |
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| 946. |
The work done in moving a particle from a point (1,1) to (2,3) in a plane and in a force field with potential `U=lambda(x+y)` is |
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Answer» Correct Answer - C |
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| 947. |
Give reason for the following: The potential energy of a freely falling object decreases progressively. Does this violate the law of conservation of energy? Why? |
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Answer» According to the law of conservation of energy, Energy can neither be created nor be destroyed. It can only be transformed from one form to another. Total energy before and after transformation remains the same. So, if the potential energy of a freely falling object decreases progressively then it is transformed into an equal amount of kinetic energy and the sum of both energies remains constant all the time. |
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| 948. |
Does potential energy of a spring decrease `//` increase when it is compressed or stretched ? |
| Answer» When a spring is compressed or stretched, potential energy energy of the spring increases in both the cases. This is because work is done by us in compression as well as stretching. | |
| 949. |
Explain with reason whether the potential energy in the following cases increases or decreases: (a) a spring is compressed, (b) a spring is stretched, (c) two dissimilar charges brought near each other, (d) a body is taken away agains the gravitational force. |
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Answer» (a) When a spring is compressed, work is done on the spring in compressing it. Therefore, P.E. increases. (b) When a spring is stretched. Work is done on the spring in stretching it. Therefore, P.E. increases. (c ) Tow dissimilar charges attract each other and come closer. Work is done by the field. Therefore, P.E. decreases. (d) Work is done by us in taking the body away against the gravitational force. Therefore, P.E. increases. (e) Air bubble rises up in water because of upthrust. Therefore, P.E. decreases. |
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| 950. |
A man weighing 60 kg climbs up a staircase carrying a load of 20 kg on his head. The stair case has 20 steps each of height 0.2 m. If he takes 10 s to climb find his powerA. 313.6 WB. 120.6 WC. 510 WD. 0 |
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Answer» Correct Answer - A Here, m =60+20=80 kg h = 20 x 0.2 = 4 m, `g = 9.8 ms^(-2)`, t = 10 s `P=W/t=(mgh)/t=(80xx9.8xx4)/10=3136/10=313.6 W` |
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