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This section includes InterviewSolutions, each offering curated multiple-choice questions to sharpen your Class 11 knowledge and support exam preparation. Choose a topic below to get started.
| 38051. |
gamma represents the ratio of two specific heats of a gas. For a given mass of the gas, the change in internal energy when the volume expands from V to 3V to constant pressure P is |
| Answer» <html><body><p>`(3PV)/((gamma - <a href="https://interviewquestions.tuteehub.com/tag/1-256655" style="font-weight:bold;" target="_blank" title="Click to know more about 1">1</a>))`<br/>`(3PV)/((gamma + 1))`<br/>`(2PV)/((gamma + 1))`<br/>`(2PV)/((gamma - 1))`</p><a href="https://interviewquestions.tuteehub.com/tag/solution-25781" style="font-weight:bold;" target="_blank" title="Click to know more about SOLUTION">SOLUTION</a> :The change in internal energy is `DeltaU = nC_(V)<a href="https://interviewquestions.tuteehub.com/tag/delta-947703" style="font-weight:bold;" target="_blank" title="Click to know more about DELTA">DELTA</a> T` <br/> But `C_(V) = (<a href="https://interviewquestions.tuteehub.com/tag/r-611811" style="font-weight:bold;" target="_blank" title="Click to know more about R">R</a> )/((gamma - 1)) :. Delta U = (nR Delta T)/((gamma - 1))` <br/> According to the ideal gas <a href="https://interviewquestions.tuteehub.com/tag/equation-974081" style="font-weight:bold;" target="_blank" title="Click to know more about EQUATION">EQUATION</a>, PV = nRT <br/> At constant pressure, `Pdelta V = nR Delta T` <br/> `:. Delta U = (PDelta V)/((gamma - 1)) = (P(3V - V))/((gamma- 1)) = (2PV)/((gamma - 1))`</body></html> | |
| 38052. |
In damped oscillatory motion a block of mass 200g is suspended to a spring of force constant 90 N//m in a medium and damping constant is 40g//s. Find time taken for its amplitude of oscillation to drop to half of its initial value |
| Answer» <html><body><p></p>Solution :Mass ,= <a href="https://interviewquestions.tuteehub.com/tag/200g-290204" style="font-weight:bold;" target="_blank" title="Click to know more about 200G">200G</a>= 0.2 kg force constant k = `90N//m` damping constant `b = 40g//s =0.04 kg//s`. <br/> `sqrt(km)= sqrt(<a href="https://interviewquestions.tuteehub.com/tag/90-341351" style="font-weight:bold;" target="_blank" title="Click to know more about 90">90</a> xx 0.2)= sqrt(18) kg//s` <br/> Here `b lt lt sqrt(km)`<br/> amplitude = `Ae^(-bt//2m)` <br/> Let amplitude is dropped to half of its initial value after the time `T_(1//2)`: <br/> Amplitude `Ae^((-bT_(1//2))/(2m))= (A)/(2) implies e^((-bt_(1//2))/(2m))= (1)/(2)` <br/> Take <a href="https://interviewquestions.tuteehub.com/tag/natural-575613" style="font-weight:bold;" target="_blank" title="Click to know more about NATURAL">NATURAL</a> logarithms on both sides <br/> `(-bT_(1//2))/(2m)= "<a href="https://interviewquestions.tuteehub.com/tag/ln-1076444" style="font-weight:bold;" target="_blank" title="Click to know more about LN">LN</a>"((1)/(2))implies T_(1//2)= ("ln"(2))/(b//2m)= 2.302 xx 0.3010xx2m//b` <br/> `T_(1//2)= 0.693 xx(2m)/(b)= 0.693xx (2 xx 0.2)/(0.04)= 6.93s`</body></html> | |
| 38053. |
A stone is thrown vertically from the ground. It reaches the maximum height of 500 m in 10 sec. After what time it will reach the ground from the maximum height reached? |
| Answer» <html><body><p>5s<br/>10s<br/>15s<br/>20s</p>Answer :<a href="https://interviewquestions.tuteehub.com/tag/b-387190" style="font-weight:bold;" target="_blank" title="Click to know more about B">B</a></body></html> | |
| 38054. |
In damped oscillatory motion a block of mass 200g is suspended to a spring of force constant 90 N//m in a medium and damping constant is 40g//s. Find time period of oscillation |
| Answer» <html><body><p></p>Solution :<a href="https://interviewquestions.tuteehub.com/tag/mass-1088425" style="font-weight:bold;" target="_blank" title="Click to know more about MASS">MASS</a> ,= 200g= 0.2 kg force constant <a href="https://interviewquestions.tuteehub.com/tag/k-527196" style="font-weight:bold;" target="_blank" title="Click to know more about K">K</a> = `90N//m` damping constant `b = 40g//s =0.04 kg//s`. <br/> `sqrt(km)= sqrt(<a href="https://interviewquestions.tuteehub.com/tag/90-341351" style="font-weight:bold;" target="_blank" title="Click to know more about 90">90</a> xx 0.2)= sqrt(18) kg//s` <br/> Here `b lt lt sqrt(km)`<br/> <a href="https://interviewquestions.tuteehub.com/tag/time-19467" style="font-weight:bold;" target="_blank" title="Click to know more about TIME">TIME</a> period `T= 2pisqrt((m)/(k))= 2pisqrt((0.2)/(90))= 0.3s`</body></html> | |
| 38055. |
A hemispherical bowl of radius R is set rotating about its axis of symmetry which is kept vertical A small block kept in the bowl rotates with the bowl without slipping on its surface If the surface of bowl is smooth and angle made by radius through the block with the vertical is 0 find the angular speed at which the bowl is rotating . |
| Answer» <html><body><p></p>Solution :Here ` OA = <a href="https://interviewquestions.tuteehub.com/tag/r-611811" style="font-weight:bold;" target="_blank" title="Click to know more about R">R</a>, <a href="https://interviewquestions.tuteehub.com/tag/angleaoc-1978297" style="font-weight:bold;" target="_blank" title="Click to know more about ANGLEAOC">ANGLEAOC</a> = theta` <br/> Block moves in a horizontal circle with centre`C` and <br/> radius` r = AC = R sin theta` <br/> `:.` In equilibrium `N cos theta = <a href="https://interviewquestions.tuteehub.com/tag/mg-1095425" style="font-weight:bold;" target="_blank" title="Click to know more about MG">MG</a>` <br/> and `N sin theta = m <a href="https://interviewquestions.tuteehub.com/tag/omega-585625" style="font-weight:bold;" target="_blank" title="Click to know more about OMEGA">OMEGA</a>^(2) (R sin theta)` <br/> `N = m omega^(2) R` <br/> From(i)`m omega^(2) R cos theta = mg` <br/> ` omega = sqrt((g)/(R cos theta))` <br/> <img src="https://d10lpgp6xz60nq.cloudfront.net/physics_images/PR_XI_V01_C03_S01_397_S01.png" width="80%"/> .</body></html> | |
| 38056. |
From a tower of height H, a particle is thrown verticially upward with a speed u. The time taken by the particle to hit the ground is n times that taken by it to reach the highest point of its path. The relation between H, u and n is |
| Answer» <html><body><p>2gH=`n^(2)u^(2)`<br/>2gH`=(n-2)^(2)u^(2)`<br/>2gH=`"nu"^(2)(n-2)` <br/>gH`=(n-2)^(2)u^(2)`</p>Solution :Time taken to <a href="https://interviewquestions.tuteehub.com/tag/reach-1178062" style="font-weight:bold;" target="_blank" title="Click to know more about REACH">REACH</a> the heighest point = `(u)/(g)` <br/> Speed on reaching the ground = `sqrt(u^(2)+2gh)` <br/> Now,<a href="https://interviewquestions.tuteehub.com/tag/v-722631" style="font-weight:bold;" target="_blank" title="Click to know more about V">V</a> = u+at <br/> or, `sqrt(u^(2)+2gh)=-u+"<a href="https://interviewquestions.tuteehub.com/tag/gt-1013864" style="font-weight:bold;" target="_blank" title="Click to know more about GT">GT</a>"` <br/> or, `t= (u+sqrt(u^(2)+2gH))/(g)` <br/> According to the <a href="https://interviewquestions.tuteehub.com/tag/equestion-2618058" style="font-weight:bold;" target="_blank" title="Click to know more about EQUESTION">EQUESTION</a>, <br/> ` (u+sqrt(u^(2)+2gH))/(g)=("nu")/(g)` <br/> or, `2gH=n(n-2)u^(2)` <br/> <img src="https://d10lpgp6xz60nq.cloudfront.net/physics_images/CHY_DMB_PHY_XI_P1_U02_C01_E07_034_S01.png" width="80%"/></body></html> | |
| 38057. |
In the Atwood's machine the system starts from rest. What is the speed and distance moved by each mass at t = 3s? |
| Answer» <html><body><p><br/></p><a href="https://interviewquestions.tuteehub.com/tag/answer-15557" style="font-weight:bold;" target="_blank" title="Click to know more about ANSWER">ANSWER</a> :`2.67ms^(-<a href="https://interviewquestions.tuteehub.com/tag/1-256655" style="font-weight:bold;" target="_blank" title="Click to know more about 1">1</a>)`, <a href="https://interviewquestions.tuteehub.com/tag/4m-318788" style="font-weight:bold;" target="_blank" title="Click to know more about 4M">4M</a></body></html> | |
| 38058. |
For bodies of regular shape and uniform mass distribution, the center of mass is at |
| Answer» <html><body><p>the corners<br/>inside the objects<br/>the <a href="https://interviewquestions.tuteehub.com/tag/point-1157106" style="font-weight:bold;" target="_blank" title="Click to know more about POINT">POINT</a> where the <a href="https://interviewquestions.tuteehub.com/tag/diagonals-950735" style="font-weight:bold;" target="_blank" title="Click to know more about DIAGONALS">DIAGONALS</a> meet<br/>the <a href="https://interviewquestions.tuteehub.com/tag/geometric-472345" style="font-weight:bold;" target="_blank" title="Click to know more about GEOMETRIC">GEOMETRIC</a> <a href="https://interviewquestions.tuteehub.com/tag/center-11455" style="font-weight:bold;" target="_blank" title="Click to know more about CENTER">CENTER</a> </p><a href="https://interviewquestions.tuteehub.com/tag/answer-15557" style="font-weight:bold;" target="_blank" title="Click to know more about ANSWER">ANSWER</a> :D</body></html> | |
| 38059. |
6.7 gm of water at 50^@C be divided in two parts such that when one part of mass .x. gm is turned into ice at 0^@C , it would release sufficient amount of heat to vaporize the other part . The (x-0.49) gm is equal to ___ gm |
| Answer» <html><body><p><br/></p><a href="https://interviewquestions.tuteehub.com/tag/answer-15557" style="font-weight:bold;" target="_blank" title="Click to know more about ANSWER">ANSWER</a> :<a href="https://interviewquestions.tuteehub.com/tag/5-319454" style="font-weight:bold;" target="_blank" title="Click to know more about 5">5</a></body></html> | |
| 38060. |
If I_(1)I_(2), and I_(3) are moments of inertia of a disc about its geometric axis, diameter and a tangent in its plane, then |
| Answer» <html><body><p>`I_(<a href="https://interviewquestions.tuteehub.com/tag/1-256655" style="font-weight:bold;" target="_blank" title="Click to know more about 1">1</a>)<a href="https://interviewquestions.tuteehub.com/tag/gt-1013864" style="font-weight:bold;" target="_blank" title="Click to know more about GT">GT</a> I_(<a href="https://interviewquestions.tuteehub.com/tag/2-283658" style="font-weight:bold;" target="_blank" title="Click to know more about 2">2</a>)gt I_(3)`<br/>`I_(3)gt I_(2)gt I_(1)`<br/>`I_(3)gt I_(1)gt I_(2)`<br/>`I_(2)gt I_(1)gt I_(3)`</p>Answer :C</body></html> | |
| 38061. |
The radii of two soap bubbles are R_1and R_2 respectively the ratio of masses of air in them will be |
| Answer» <html><body><p><<a href="https://interviewquestions.tuteehub.com/tag/p-588962" style="font-weight:bold;" target="_blank" title="Click to know more about P">P</a>>`(R_(1)^(<a href="https://interviewquestions.tuteehub.com/tag/3-301577" style="font-weight:bold;" target="_blank" title="Click to know more about 3">3</a>))/( R_(2)^(3))`<br/>`(R_(2)^(3))/( R_(1)^(3))`<br/>`((P+ (4T )/ (R_1) )/(P +(4T)/(R_2)))(R_(1)^(3))/( R_(2)^(3))`<br/>`((P+ (4T )/( R_2) )/(P +(4T)/(R_1)))(R_(2)^(3))/( R_(1)^(3))`</p><a href="https://interviewquestions.tuteehub.com/tag/answer-15557" style="font-weight:bold;" target="_blank" title="Click to know more about ANSWER">ANSWER</a> :<a href="https://interviewquestions.tuteehub.com/tag/c-7168" style="font-weight:bold;" target="_blank" title="Click to know more about C">C</a></body></html> | |
| 38062. |
If the diameter of the earth becomes half its present value but its average density remains unchanged then how would the weight of an object on the surface of the earth be affected? |
| Answer» <html><body><p></p>Solution :<a href="https://interviewquestions.tuteehub.com/tag/g-1003017" style="font-weight:bold;" target="_blank" title="Click to know more about G">G</a> = `GM/R^2` = `G/R^2 4/3 piR^3rho` = `4/3 piGR <a href="https://interviewquestions.tuteehub.com/tag/rho-623364" style="font-weight:bold;" target="_blank" title="Click to know more about RHO">RHO</a>` <br/> g = `4/3 piG (R/2) rho` = g/2 . Hence the weight of the <a href="https://interviewquestions.tuteehub.com/tag/object-11416" style="font-weight:bold;" target="_blank" title="Click to know more about OBJECT">OBJECT</a> will be <a href="https://interviewquestions.tuteehub.com/tag/halved-2104643" style="font-weight:bold;" target="_blank" title="Click to know more about HALVED">HALVED</a>.</body></html> | |
| 38063. |
If the linear density of a rod of length L varies as lambda=(kx^(2))/(L) where k is a constant and x is the distance of any point from one end, then find the distance of centre of mass from the end at x=0. |
| Answer» <html><body><p></p>Solution :Let the x-axis be along the length of the rod and origin at one of its ends. As rod is along x-axis, for all points on it y and z co-ordinates are zero. <br/> <img src="https://doubtnut-static.s.llnwi.net/static/physics_images/AKS_NEO_CAO_PHY_XI_V01_B_C07_SLV_023_S01.png" width="80%"/> <br/> Centre of mass will be on the rod. Now consider an element of rod of length dx at a distance x from the origin, then `<a href="https://interviewquestions.tuteehub.com/tag/dm-432223" style="font-weight:bold;" target="_blank" title="Click to know more about DM">DM</a>=lambda` dx`=(<a href="https://interviewquestions.tuteehub.com/tag/kx-1064991" style="font-weight:bold;" target="_blank" title="Click to know more about KX">KX</a>^(2))/(L)dx` <br/> so, `X_(CM)=(int_(0)^(L)xdm)/(int_(0)^(L)dm)=(int_(0)^(L)x(kx^(2))/(L)dx)/(int_(0)^(L)(kx^(2))/(L)dx)=(int_(0)^(L)x^(3)dx)/(int_(0)^(L)x^(2)dx)=((L^(<a href="https://interviewquestions.tuteehub.com/tag/4-311707" style="font-weight:bold;" target="_blank" title="Click to know more about 4">4</a>))/(4))/((L^(3))/(3))=(<a href="https://interviewquestions.tuteehub.com/tag/3l-310759" style="font-weight:bold;" target="_blank" title="Click to know more about 3L">3L</a>)/(4)`</body></html> | |
| 38064. |
When a disc rotates with unform angular velocity, which of the following is not true ? |
| Answer» <html><body><p>The sense of <a href="https://interviewquestions.tuteehub.com/tag/rotation-11281" style="font-weight:bold;" target="_blank" title="Click to know more about ROTATION">ROTATION</a> remains same <br/>The <a href="https://interviewquestions.tuteehub.com/tag/orientation-587745" style="font-weight:bold;" target="_blank" title="Click to know more about ORIENTATION">ORIENTATION</a> of the <a href="https://interviewquestions.tuteehub.com/tag/axis-889931" style="font-weight:bold;" target="_blank" title="Click to know more about AXIS">AXIS</a> of rotation remains same <br/>The speed of rotation is non-zero and remains same <br/>The sngular <a href="https://interviewquestions.tuteehub.com/tag/acceleration-13745" style="font-weight:bold;" target="_blank" title="Click to know more about ACCELERATION">ACCELERATION</a> is non-zero and remains same </p>Answer :D</body></html> | |
| 38065. |
An ideal gas undergoes a quasistatic, reversible process in which its molar heat capacity C remains constant. It during this process the relation the pressure p and volume V is given by pV^n=constant, then n is given by (here C_p and C_v are molar specific heat at constant pressure and constant volume respectively). |
| Answer» <html><body><p>`n=C_p/C_v`<br/>`n=(<a href="https://interviewquestions.tuteehub.com/tag/c-7168" style="font-weight:bold;" target="_blank" title="Click to know more about C">C</a>-C_p)/(C-C_v)`<br/>`n=(C_p-C)/(C-C_v)`<br/>`n=(C-C_v)/(C-C_p)`</p>Solution :Here `<a href="https://interviewquestions.tuteehub.com/tag/pv-593601" style="font-weight:bold;" target="_blank" title="Click to know more about PV">PV</a>^n=k` (constant)…….(<a href="https://interviewquestions.tuteehub.com/tag/1-256655" style="font-weight:bold;" target="_blank" title="Click to know more about 1">1</a>) <br/> For 1 mol of ideal gas <br/> `pV=RT`……..(2) <br/> Dividing (1) by (2) we get `V^(n-1) T=k/R` <br/> `<a href="https://interviewquestions.tuteehub.com/tag/therefore-706901" style="font-weight:bold;" target="_blank" title="Click to know more about THEREFORE">THEREFORE</a>((dV)/(dT))=V/((n-1)T)=V/((1-n)T)`<br/>According to first law of <a href="https://interviewquestions.tuteehub.com/tag/thermodynamics-12722" style="font-weight:bold;" target="_blank" title="Click to know more about THERMODYNAMICS">THERMODYNAMICS</a> <br/> `dQ=C_vdT+pdV`<br/> `therefore(dQ)/(dT)=C_V+p((dV)/(dT))=C_v+(pV)/((1-n)T)=C_v+R/(1-n)`<br/>Hence thermal capacity, `C=C_v+R/(1-n)` <br/> or,`1-n=R/(C-C_v)`<br/> or,`n=1-R/(C-C_v)=(C-(C_v-R))/(C-V_v)=(C-C_p)/(C-C_v)`[`because C_p-C_v=R`]</body></html> | |
| 38066. |
A bucket filled with water weighing 20 kg is raised from a well of depth 20m. If the lineat density of the rope is 0.2 kg per meter, the amount of work done is, (g=10 ms^(-2)) |
| Answer» <html><body><p><a href="https://interviewquestions.tuteehub.com/tag/4000-315574" style="font-weight:bold;" target="_blank" title="Click to know more about 4000">4000</a> <a href="https://interviewquestions.tuteehub.com/tag/j-520843" style="font-weight:bold;" target="_blank" title="Click to know more about J">J</a><br/>4040 J<br/>4400 J <br/><a href="https://interviewquestions.tuteehub.com/tag/4200-1876189" style="font-weight:bold;" target="_blank" title="Click to know more about 4200">4200</a> J</p>Answer :<a href="https://interviewquestions.tuteehub.com/tag/c-7168" style="font-weight:bold;" target="_blank" title="Click to know more about C">C</a></body></html> | |
| 38067. |
A light particle moving horizontally with a speed of 12 m/s strikes a very heavy block moving in the same direction at 10 m/s. The collision is one dimensional and elastic. After the collision, the particle will be |
| Answer» <html><body><p>move at 2 m/s in its <a href="https://interviewquestions.tuteehub.com/tag/original-1139424" style="font-weight:bold;" target="_blank" title="Click to know more about ORIGINAL">ORIGINAL</a> direction <br/>move at 8 m/s in its original direction <br/>move at 8 m/s opposite to its original direction <br/>move at 12 m/s opposite to its original direction </p>Solution :Here, `u_1 = 12 m//s , u_2 = <a href="https://interviewquestions.tuteehub.com/tag/10-261113" style="font-weight:bold;" target="_blank" title="Click to know more about 10">10</a> m//s` <br/> <a href="https://interviewquestions.tuteehub.com/tag/let-11597" style="font-weight:bold;" target="_blank" title="Click to know more about LET">LET</a> `v_1` be velocity of the light particle after <a href="https://interviewquestions.tuteehub.com/tag/collision-922060" style="font-weight:bold;" target="_blank" title="Click to know more about COLLISION">COLLISION</a>. <br/> `v_1 = ((m_1 -m_2)u_1)/(m_1 + m_2) + (2m_2 u_2)/(m_1 + m_2) "" ……..(i)` <br/> Given : `m_1 < < m_2` <br/> So, `m_1` can be ignored compared to `m_2`. <br/> From equation (i) we get , `v_1 = - u_1 + 2u_2` <br/> Substituting the <a href="https://interviewquestions.tuteehub.com/tag/values-25920" style="font-weight:bold;" target="_blank" title="Click to know more about VALUES">VALUES</a> , we get <br/> `v_1 = -12 m//s + 2(10 m//s) = 8 m//s` in its original direction.</body></html> | |
| 38068. |
In the figure, a smooth pulley of negligible weight is suspended by a spring balance. Weights of 1 kg and 5kg are attached to the opposite ends of a string passing over the pulley and move with an acceleration due of gravity. During their motion, the spring balance reads a weight of |
| Answer» <html><body><p><a href="https://interviewquestions.tuteehub.com/tag/6-327005" style="font-weight:bold;" target="_blank" title="Click to know more about 6">6</a> kg<br/>less than 6 kg <br/>more than 6 kg <br/><a href="https://interviewquestions.tuteehub.com/tag/may-557248" style="font-weight:bold;" target="_blank" title="Click to know more about MAY">MAY</a> be more or less than 6kg</p>Solution :`m_(<a href="https://interviewquestions.tuteehub.com/tag/1-256655" style="font-weight:bold;" target="_blank" title="Click to know more about 1">1</a>)g-T=m_(1)a, T-m_(2)g=m_(2)a, T^(1)=<a href="https://interviewquestions.tuteehub.com/tag/2t-301164" style="font-weight:bold;" target="_blank" title="Click to know more about 2T">2T</a>`</body></html> | |
| 38069. |
The fire fighters attach brass jets at the end of water pipes. Why? |
| Answer» <html><body><p></p><a href="https://interviewquestions.tuteehub.com/tag/solution-25781" style="font-weight:bold;" target="_blank" title="Click to know more about SOLUTION">SOLUTION</a> :To Increase the <a href="https://interviewquestions.tuteehub.com/tag/velocity-1444512" style="font-weight:bold;" target="_blank" title="Click to know more about VELOCITY">VELOCITY</a> of water issuing out of the nozzle as the area a of the <a href="https://interviewquestions.tuteehub.com/tag/tip-1420809" style="font-weight:bold;" target="_blank" title="Click to know more about TIP">TIP</a> is very <a href="https://interviewquestions.tuteehub.com/tag/small-1212368" style="font-weight:bold;" target="_blank" title="Click to know more about SMALL">SMALL</a>, increases in agreement with the equation of continuity ay a constant</body></html> | |
| 38070. |
For a progressive wave, y = 5 sin (0.01 x - 2t) (where x and y are in cm and t is in s). What will be its speed od propagation ? |
| Answer» <html><body><p></p><a href="https://interviewquestions.tuteehub.com/tag/solution-25781" style="font-weight:bold;" target="_blank" title="Click to know more about SOLUTION">SOLUTION</a> :Comparing `y =5 sin (0.01x -2t)` with ` y = A sin (kx - omega t )` <br/> We get `k =0.01 rad//cm, omega =2 rad//s` <br/> Now, <a href="https://interviewquestions.tuteehub.com/tag/wave-22399" style="font-weight:bold;" target="_blank" title="Click to know more about WAVE">WAVE</a> <a href="https://interviewquestions.tuteehub.com/tag/velocity-1444512" style="font-weight:bold;" target="_blank" title="Click to know more about VELOCITY">VELOCITY</a> is `<a href="https://interviewquestions.tuteehub.com/tag/v-722631" style="font-weight:bold;" target="_blank" title="Click to know more about V">V</a> = (omega )/(k) = (2)/(0.01) =<a href="https://interviewquestions.tuteehub.com/tag/200-288914" style="font-weight:bold;" target="_blank" title="Click to know more about 200">200</a> cm//s`</body></html> | |
| 38071. |
The force exerted by the floor of an elevator on the foot of a person standing there is more than the weight of the person if the elevator is (a) going up and slowing down(b) going up and speeding up(c ) going down and slowing down(d) going down and speeding up |
| Answer» <html><body><p><a href="https://interviewquestions.tuteehub.com/tag/b-387190" style="font-weight:bold;" target="_blank" title="Click to know more about B">B</a> and <a href="https://interviewquestions.tuteehub.com/tag/c-7168" style="font-weight:bold;" target="_blank" title="Click to know more about C">C</a> are <a href="https://interviewquestions.tuteehub.com/tag/correct-409949" style="font-weight:bold;" target="_blank" title="Click to know more about CORRECT">CORRECT</a> <br/>a and b are correct <br/>b and d are correct <br/>d only correct </p><a href="https://interviewquestions.tuteehub.com/tag/answer-15557" style="font-weight:bold;" target="_blank" title="Click to know more about ANSWER">ANSWER</a> :A</body></html> | |
| 38072. |
Certain volume of a liquid is taken in a Jong glass tube and its temperature is increased at a uniform rate, the rate of increase in the length of the liquid depends on (a) length of the liquid coloumn (b) area of cross section of the glass tube (c) coefficient of expansion of glass |
| Answer» <html><body><p>only (a) is <a href="https://interviewquestions.tuteehub.com/tag/correct-409949" style="font-weight:bold;" target="_blank" title="Click to know more about CORRECT">CORRECT</a> <br/>(a) & (<a href="https://interviewquestions.tuteehub.com/tag/b-387190" style="font-weight:bold;" target="_blank" title="Click to know more about B">B</a>) are correct <br/>(b) & (<a href="https://interviewquestions.tuteehub.com/tag/c-7168" style="font-weight:bold;" target="_blank" title="Click to know more about C">C</a>) are correct <br/>(a) (b) & (c) are correct </p><a href="https://interviewquestions.tuteehub.com/tag/answer-15557" style="font-weight:bold;" target="_blank" title="Click to know more about ANSWER">ANSWER</a> :D</body></html> | |
| 38073. |
A uniform wire of length 6.28 cm is bent in the form of a circle. The shift in its centre of mass is |
| Answer» <html><body><p><a href="https://interviewquestions.tuteehub.com/tag/1-256655" style="font-weight:bold;" target="_blank" title="Click to know more about 1">1</a> <a href="https://interviewquestions.tuteehub.com/tag/cm-919986" style="font-weight:bold;" target="_blank" title="Click to know more about CM">CM</a><br/>2 cm<br/>4 cm<br/>`3.14 cm`</p><a href="https://interviewquestions.tuteehub.com/tag/answer-15557" style="font-weight:bold;" target="_blank" title="Click to know more about ANSWER">ANSWER</a> :A</body></html> | |
| 38074. |
The amount of heat energy required to raise the temperature of 1 g of helium at NTP, from T_(1)K to T_(2)K is |
| Answer» <html><body><p>`(<a href="https://interviewquestions.tuteehub.com/tag/3-301577" style="font-weight:bold;" target="_blank" title="Click to know more about 3">3</a>)/(<a href="https://interviewquestions.tuteehub.com/tag/2-283658" style="font-weight:bold;" target="_blank" title="Click to know more about 2">2</a>)N_(a)k_(B)(T_(2)-T_(<a href="https://interviewquestions.tuteehub.com/tag/1-256655" style="font-weight:bold;" target="_blank" title="Click to know more about 1">1</a>))`<br/>`(3)/(4)N_(a)k_(B)(T_(2)-T_(1))`<br/>`(3)/(4)N_(a)k_(B)((T_(2))/(T_(1)))`<br/>`(3)/(8)N_(a)K_(B)(T_(2)-T_(1))`</p><a href="https://interviewquestions.tuteehub.com/tag/answer-15557" style="font-weight:bold;" target="_blank" title="Click to know more about ANSWER">ANSWER</a> :D</body></html> | |
| 38075. |
Optical fibers are based on |
| Answer» <html><body><p><a href="https://interviewquestions.tuteehub.com/tag/total-711110" style="font-weight:bold;" target="_blank" title="Click to know more about TOTAL">TOTAL</a> <a href="https://interviewquestions.tuteehub.com/tag/internal-517481" style="font-weight:bold;" target="_blank" title="Click to know more about INTERNAL">INTERNAL</a> reflection<br/>refraction<br/>dispersion<br/>none of these</p>Answer :(a)</body></html> | |
| 38076. |
The displacement of a particle is given by y (t) =2t ^(2) +5m.Hence its velocity at the end of 6 sec. will be ...... m/s. |
| Answer» <html><body><p>77<br/><a href="https://interviewquestions.tuteehub.com/tag/4-311707" style="font-weight:bold;" target="_blank" title="Click to know more about 4">4</a><br/>0<br/>24</p>Solution :`y = 2 t ^(2) + 5` <br/> `v = <a href="https://interviewquestions.tuteehub.com/tag/4t-319060" style="font-weight:bold;" target="_blank" title="Click to know more about 4T">4T</a>` <br/> `therefore t = <a href="https://interviewquestions.tuteehub.com/tag/6-327005" style="font-weight:bold;" target="_blank" title="Click to know more about 6">6</a> <a href="https://interviewquestions.tuteehub.com/tag/sec-1197209" style="font-weight:bold;" target="_blank" title="Click to know more about SEC">SEC</a>.` speed <br/> `therefore v = 4 xx 6 = 24m//s`</body></html> | |
| 38077. |
A metallic sphere of mass M falls through glycerine with a terminal velocity v. If we drop a ball of mass 8M of same metal into a column of glycerine, the terminal velocity of ball will be |
| Answer» <html><body><p><a href="https://interviewquestions.tuteehub.com/tag/2v-300496" style="font-weight:bold;" target="_blank" title="Click to know more about 2V">2V</a> <br/>4v<br/>8v<br/>16v</p>Solution :`M=(4)/(3)pir^(3)<a href="https://interviewquestions.tuteehub.com/tag/rho-623364" style="font-weight:bold;" target="_blank" title="Click to know more about RHO">RHO</a> and <a href="https://interviewquestions.tuteehub.com/tag/8m-340185" style="font-weight:bold;" target="_blank" title="Click to know more about 8M">8M</a> =(4)/(3)piR^(3)rho`, <br/> `therefore R^(3)=8r^(3)` or `R=<a href="https://interviewquestions.tuteehub.com/tag/2r-300472" style="font-weight:bold;" target="_blank" title="Click to know more about 2R">2R</a>` <br/> As `v prop r^(2)` <br/> `therefore (v')/(v)=((R)/(r))^(2)=((2r)/(r))^(2)=4` or `v'=4v`</body></html> | |
| 38078. |
Which of the following properties are suitable for a cooking utensil ? |
| Answer» <html><body><p><a href="https://interviewquestions.tuteehub.com/tag/high-479925" style="font-weight:bold;" target="_blank" title="Click to know more about HIGH">HIGH</a> specificheat <br/><a href="https://interviewquestions.tuteehub.com/tag/low-537644" style="font-weight:bold;" target="_blank" title="Click to know more about LOW">LOW</a> <a href="https://interviewquestions.tuteehub.com/tag/specific-1220917" style="font-weight:bold;" target="_blank" title="Click to know more about SPECIFIC">SPECIFIC</a> heat <br/>High <a href="https://interviewquestions.tuteehub.com/tag/conductivity-17908" style="font-weight:bold;" target="_blank" title="Click to know more about CONDUCTIVITY">CONDUCTIVITY</a> <br/>Low conductivity </p>Answer :B::<a href="https://interviewquestions.tuteehub.com/tag/c-7168" style="font-weight:bold;" target="_blank" title="Click to know more about C">C</a></body></html> | |
| 38079. |
A particle of mass m is suspended from the ceiling through a string of length L. The particle moves in a horizontal circle of radius r. The speed of the particle is |
| Answer» <html><body><p>`(rg)/(sqrt(L^(2)-<a href="https://interviewquestions.tuteehub.com/tag/r-611811" style="font-weight:bold;" target="_blank" title="Click to know more about R">R</a>^(2)))`<br/>`(rsqrtg)/((L^(2)-r^(2))^((<a href="https://interviewquestions.tuteehub.com/tag/1-256655" style="font-weight:bold;" target="_blank" title="Click to know more about 1">1</a>)/(4)))`<br/>`(rsqrtg)/((L^(2)-r^(2))^((1)/(2)))`<br/>`(<a href="https://interviewquestions.tuteehub.com/tag/mgl-2176708" style="font-weight:bold;" target="_blank" title="Click to know more about MGL">MGL</a>)/((L^(2)-r^(2))^((1)/(2)))`</p>Solution :`(F)/(r )=(mg)/(sqrt(L^(2)-r^(2)))`</body></html> | |
| 38080. |
A sphere is rolling down on a sufficiently rough inclined surface. |
| Answer» <html><body><p>Direction of friction is up the plane <br/>Total mechanical energy of the sphere remains constant <br/>Angular momentum of sphere remains conserved about instantaneous point of <a href="https://interviewquestions.tuteehub.com/tag/contact-25916" style="font-weight:bold;" target="_blank" title="Click to know more about CONTACT">CONTACT</a><br/></p>Solution :When sphere <a href="https://interviewquestions.tuteehub.com/tag/tries-3232458" style="font-weight:bold;" target="_blank" title="Click to know more about TRIES">TRIES</a> to slip down the plane <a href="https://interviewquestions.tuteehub.com/tag/due-433472" style="font-weight:bold;" target="_blank" title="Click to know more about DUE">DUE</a> to component of its weight along the same plane then friction acts upon the plane to oppose the <a href="https://interviewquestions.tuteehub.com/tag/possible-592355" style="font-weight:bold;" target="_blank" title="Click to know more about POSSIBLE">POSSIBLE</a> relative motion at the contact. Friction acting this way provides necessary torque about the centre for rotational motion so that sphere may perform pure rolling. In case of pure rolling instantaneous velocity of the point of contact is zero and this is the point where friction acts hence work done by the friction remains zero and total mechanical energy remains conserved because there is no heat loss due to friction. Note that there is torque of weight of the sphere about point of contact hence angular momentum of the sphere about point of contact cannot remain constant. So options (a), (b) and (c) are <a href="https://interviewquestions.tuteehub.com/tag/correct-409949" style="font-weight:bold;" target="_blank" title="Click to know more about CORRECT">CORRECT</a>.</body></html> | |
| 38081. |
A tuning fork is used to produce resonance in a glass tube . The length of the air column in this tube can be adjusled by a varlable pision. At room temperature of 27^(@) C, two successlve resonances are produced at20 cm and 73 cm of column length .If the frequency of the tuning fork is 320 Hz, the velocity of sound in air at 27^(@) C is |
| Answer» <html><body><p><a href="https://interviewquestions.tuteehub.com/tag/350-308849" style="font-weight:bold;" target="_blank" title="Click to know more about 350">350</a> m/s<br/>339m/s<br/>330 m/s<br/>300 m/s</p>Solution :`(2n= l)(lambda)/(4) = 20`<br/>`(2 n + 1) (lambda)/(4) = <a href="https://interviewquestions.tuteehub.com/tag/73-334616" style="font-weight:bold;" target="_blank" title="Click to know more about 73">73</a>`<br/>Subtracting <a href="https://interviewquestions.tuteehub.com/tag/equation-974081" style="font-weight:bold;" target="_blank" title="Click to know more about EQUATION">EQUATION</a> (2) from equation (1) . We get<br/>`(2 n + 1 - 2 n + 1) (lambda)/(4) = (73 - 20)`<br/>or,`(lambda)/(2) = <a href="https://interviewquestions.tuteehub.com/tag/53-324827" style="font-weight:bold;" target="_blank" title="Click to know more about 53">53</a> or, lambda = 106 cm = (106)/(<a href="https://interviewquestions.tuteehub.com/tag/100-263808" style="font-weight:bold;" target="_blank" title="Click to know more about 100">100</a>) m`<br/>`:. (V) = nlambda = 320 xx (106)/(100) = 339 . 2 m//2`</body></html> | |
| 38082. |
Smooth block is released at rest on a 45^@incline and then slides a distance d. If the time taken of slide on rough incline is n times as large as that to slide than on a smooth incline. Show that coefficient of friction. mu = (1- (1)/(n^2)) |
| Answer» <html><body><p></p>Solution :When there is no friction, the block slides down the inclined plane with <a href="https://interviewquestions.tuteehub.com/tag/acceleration-13745" style="font-weight:bold;" target="_blank" title="Click to know more about ACCELERATION">ACCELERATION</a>. <br/>` a = g <a href="https://interviewquestions.tuteehub.com/tag/sin-1208945" style="font-weight:bold;" target="_blank" title="Click to know more about SIN">SIN</a> theta `<br/>when there is friction, the downward acceleration of the block is <br/> ` a. =g (sin theta - mu <a href="https://interviewquestions.tuteehub.com/tag/cos-935872" style="font-weight:bold;" target="_blank" title="Click to know more about COS">COS</a> theta)` <br/>As the block slides a distance d in each case so <br/> ` d = 1/2 at^2 = 1/2 a.t.^2` <br/>`(a)/(a.) = (t.^2)/(t^2) = ((nt)^2)/(t^2) = n^2 " or"(g sin theta)/(g (sin theta -mu cos theta) )=n^2` <br/>Solving , we get (using ` theta =<a href="https://interviewquestions.tuteehub.com/tag/45-316951" style="font-weight:bold;" target="_blank" title="Click to know more about 45">45</a>^@` ) <br/>`mu = 1 - (1)/(n^2)`</body></html> | |
| 38083. |
A ball of mass 'm' approaches a wall of mass 'M' (gt gtm) with speed 4m/s along the normal to the wall. The speed of wall is 1m/s towards the ball. The speed of the ball after an elastic collision with the wall is |
| Answer» <html><body><p>5m/s <a href="https://interviewquestions.tuteehub.com/tag/away-386546" style="font-weight:bold;" target="_blank" title="Click to know more about AWAY">AWAY</a> from the wall<br/>9m/s away from the wall<br/>6m/s away from the wall<br/>3m/s away from the wall</p>Answer :<a href="https://interviewquestions.tuteehub.com/tag/c-7168" style="font-weight:bold;" target="_blank" title="Click to know more about C">C</a></body></html> | |
| 38084. |
Intensity of gravitational field inside the hollow spherical shell is |
| Answer» <html><body><p><a href="https://interviewquestions.tuteehub.com/tag/variable-1442996" style="font-weight:bold;" target="_blank" title="Click to know more about VARIABLE">VARIABLE</a> <br/>minimum<br/>maximum<br/>zero</p>Answer :D</body></html> | |
| 38085. |
In the question number 36, at which distance a weight may be hung along the rod, in order to produce equal strains in both the wires? |
| Answer» <html><body><p>`(4)/(<a href="https://interviewquestions.tuteehub.com/tag/3-301577" style="font-weight:bold;" target="_blank" title="Click to know more about 3">3</a>)m`from steel wire<br/>`(4)/(3)m`from brass wire<br/>1 m from steel wire <br/>`(1)/(4)m`from brass wire</p>Solution :As Strain `= ("Srtress")/("Young's Modulus")` <a href="https://interviewquestions.tuteehub.com/tag/ltbr-2804405" style="font-weight:bold;" target="_blank" title="Click to know more about LTBR">LTBR</a> `therefore` Strainin stell wire `= (T_(S))/(A_(S)Y_(S))` <br/>For equal strain in both the wire,<br/> `(T_(S))/(A_(S)Y_(S)) = (T_(<a href="https://interviewquestions.tuteehub.com/tag/b-387190" style="font-weight:bold;" target="_blank" title="Click to know more about B">B</a>))/(A_(B)Y_(B))`<br/> `therefore (T_(S))/(T_(B))=(A_(S))/(A_(B)) (Y_(S))/(Y_(B)) = (0.1 cm^(2))/(0.2 cm^(2)) xx (2xx <a href="https://interviewquestions.tuteehub.com/tag/10-261113" style="font-weight:bold;" target="_blank" title="Click to know more about 10">10</a>^(<a href="https://interviewquestions.tuteehub.com/tag/11-267621" style="font-weight:bold;" target="_blank" title="Click to know more about 11">11</a>)N m^(-2))/(1 xx 10^(11) N m^(-2))= 1 .......(i)` <br/> For therotational equlibrium of the rod, <br/> `T_(S)x = T_(B) (2-x)` <br/>`(2-x)/(x) = (T_(S))/(T_(B))= 1""["Using (i)"]` <br/> `2 - x = x or x=1m `</body></html> | |
| 38086. |
If you place your hands on a wooden and a metal chair at the same temperature,lower than your body temperature, which of them do you find hotter? |
| Answer» <html><body><p></p><a href="https://interviewquestions.tuteehub.com/tag/solution-25781" style="font-weight:bold;" target="_blank" title="Click to know more about SOLUTION">SOLUTION</a> :The <a href="https://interviewquestions.tuteehub.com/tag/wooden-744658" style="font-weight:bold;" target="_blank" title="Click to know more about WOODEN">WOODEN</a> <a href="https://interviewquestions.tuteehub.com/tag/chair-415628" style="font-weight:bold;" target="_blank" title="Click to know more about CHAIR">CHAIR</a>.</body></html> | |
| 38087. |
Let the speed of the planet at the perihelion P be V_pand the sun-planet distance SP be r_p Relates (r_p,V_P )to the corresponding quantities at the aphelion (r_(A),V_A)Will the planet take equal times to tranverse BAC and CPB ? |
| Answer» <html><body><p></p>Solution :<img src="https://doubtnut-static.s.llnwi.net/static/physics_images/AKS_NEO_CAO_PHY_XI_V01_MP2_C09_SLV_004_S01.png" width="80%"/> <br/> According to <a href="https://interviewquestions.tuteehub.com/tag/law-184" style="font-weight:bold;" target="_blank" title="Click to know more about LAW">LAW</a> of <a href="https://interviewquestions.tuteehub.com/tag/conservation-929810" style="font-weight:bold;" target="_blank" title="Click to know more about CONSERVATION">CONSERVATION</a> of angular <a href="https://interviewquestions.tuteehub.com/tag/momentum-1100588" style="font-weight:bold;" target="_blank" title="Click to know more about MOMENTUM">MOMENTUM</a>. Angular momentum of the planet at P = Angular momentum of the planet at A<br/> `implies mV_pr_(p) = mV_(A)r_(A) " (or) " V_p/V_A=r_(p)/r_(p)` Since `r_A gt r_p` so `V_p gt V_A` <br/> Here area SBAC is greater than the area SCPB. <br/> According to Kepler.s second law, as areal velocity of the planet , is constant around the sun. i.e. equal areas areas are swept on equal times, hence the planet will take longer time to traverse BAC then <a href="https://interviewquestions.tuteehub.com/tag/cpb-410333" style="font-weight:bold;" target="_blank" title="Click to know more about CPB">CPB</a>.</body></html> | |
| 38088. |
4.0 g of a gas occupies 22.4 litres at NTP. The specific heat capacity of the gas at constant volume is 5.0 "JK"^(-1) "mol"^(-1). If the speed of sound in this gas at NTP is 952 "ms"^(-1) , then the heat capacity at constant pressure is _____ (R=8.3 "JK"^(1) "mol"^(-1)) |
| Answer» <html><body><p>`8.5 "JK"^(-1) "<a href="https://interviewquestions.tuteehub.com/tag/mol-1100196" style="font-weight:bold;" target="_blank" title="Click to know more about MOL">MOL</a>"^(-1)` <br/>`8.0 "JK"^(-1) "mol"^(-1)` <br/>`7.5 "JK"^(-1) "mol"^(-1)` <br/>`7.0 "JK"^(-1) "mol"^(-1)` </p>Solution :<a href="https://interviewquestions.tuteehub.com/tag/velocity-1444512" style="font-weight:bold;" target="_blank" title="Click to know more about VELOCITY">VELOCITY</a> of <a href="https://interviewquestions.tuteehub.com/tag/sound-648690" style="font-weight:bold;" target="_blank" title="Click to know more about SOUND">SOUND</a> , <br/> `v=sqrt((gammaRT)/M)` <br/> `therefore gamma=(mv^2)/(<a href="https://interviewquestions.tuteehub.com/tag/rt-615359" style="font-weight:bold;" target="_blank" title="Click to know more about RT">RT</a>)` <br/> `=(4xx10^(-3)xx(952)^2)/(8.3xx273)` <br/> `=3625.2/2265.9` <br/> `therefore gamma=1.599` <br/> `therefore gamma approx` 1.6 <br/>`therefore` The <a href="https://interviewquestions.tuteehub.com/tag/heat-21102" style="font-weight:bold;" target="_blank" title="Click to know more about HEAT">HEAT</a> capacity at constant pressure , <br/> `C_P=gammaC_V=1.6xx5.0=8.0 "JK"^(1) "mol"^(-1)`</body></html> | |
| 38089. |
What are the conditions necessary for a satellite to appear stationary. |
| Answer» <html><body><p></p>Solution :`implies ` The conditions <a href="https://interviewquestions.tuteehub.com/tag/necessary-1112675" style="font-weight:bold;" target="_blank" title="Click to know more about NECESSARY">NECESSARY</a> for a satellite to appear stationary are : <br/> 1)Satellite having the circular orbit is in the equatorial plane of the earth. <br/> 2) Its <a href="https://interviewquestions.tuteehub.com/tag/orbital-1138169" style="font-weight:bold;" target="_blank" title="Click to know more about ORBITAL">ORBITAL</a> time period of revolution is 24 hours. <br/> 3)The satellite must <a href="https://interviewquestions.tuteehub.com/tag/travel-1426745" style="font-weight:bold;" target="_blank" title="Click to know more about TRAVEL">TRAVEL</a> eastward (<a href="https://interviewquestions.tuteehub.com/tag/west-1451899" style="font-weight:bold;" target="_blank" title="Click to know more about WEST">WEST</a> - east) at the same rotation speed as the earth.<br/> 4) It should be at a height nearly 36000 km above the equator of earth. The inclination of the orbit must be <a href="https://interviewquestions.tuteehub.com/tag/zero-751093" style="font-weight:bold;" target="_blank" title="Click to know more about ZERO">ZERO</a>.</body></html> | |
| 38090. |
The sciences which deal with …….. Are called physical sciences. |
| Answer» <html><body><p><br/></p><a href="https://interviewquestions.tuteehub.com/tag/answer-15557" style="font-weight:bold;" target="_blank" title="Click to know more about ANSWER">ANSWER</a> :<a href="https://interviewquestions.tuteehub.com/tag/non-580180" style="font-weight:bold;" target="_blank" title="Click to know more about NON">NON</a> - <a href="https://interviewquestions.tuteehub.com/tag/living-1076292" style="font-weight:bold;" target="_blank" title="Click to know more about LIVING">LIVING</a> thigs.</body></html> | |
| 38091. |
A man swims from a point A on the bank of a river if width 100 m. When he swims perpendicular to the water current, he reaches the other bank 50 m downstream. The angle to the bank at which he should swim, to reach the directly opposite point B on the other bank is. . |
| Answer» <html><body><p>`10^@` upstream<br/>`20^@` upstream<br/>`<a href="https://interviewquestions.tuteehub.com/tag/30-304807" style="font-weight:bold;" target="_blank" title="Click to know more about 30">30</a>^@` upstream<br/>`<a href="https://interviewquestions.tuteehub.com/tag/60-328817" style="font-weight:bold;" target="_blank" title="Click to know more about 60">60</a>^@` upstream</p>Solution :(d) <a href="https://interviewquestions.tuteehub.com/tag/refer-621103" style="font-weight:bold;" target="_blank" title="Click to know more about REFER">REFER</a> to (Fig. S5.68) (a), <br/> `tan <a href="https://interviewquestions.tuteehub.com/tag/theta-1412757" style="font-weight:bold;" target="_blank" title="Click to know more about THETA">THETA</a> = (v_w)/(v_m) = (50)/(100) = (1)/(2) or v_m = 2v_w` <br/> Refer to (Fig .S5.68) (<a href="https://interviewquestions.tuteehub.com/tag/b-387190" style="font-weight:bold;" target="_blank" title="Click to know more about B">B</a>), <br/> `sin prop = (v_w)/(v_m) = (v_w)/(2 v_w) = (1)/(2)` <br/> or `theta = 30^@` <br/> So, it is `60^@` upstream. <br/> (a) <img src="https://d10lpgp6xz60nq.cloudfront.net/physics_images/BMS_V01_C05_E01_106_S01.png" width="80%"/>,</body></html> | |
| 38092. |
A block of mass m resting on a smooth horizontal plane starts moving due to a constant force F=mg//3 of constantmagnitude. In the process of its rectilinear motion the angle theta varies as theta=lamdas, where lamda is constant and s is the distance traversed by the block from its intial position. Find the velocity of the block as a function of the angle theta. Also show that the block will never lose contact with the ground. |
| Answer» <html><body><p></p>Solution :<img src="https://d10lpgp6xz60nq.cloudfront.net/physics_images/CPS_V01_C06_S01_046_S01.png" width="80%"/> <br/> `x-`direction`:` `Fcostheta=ma` <br/> `a=(F)/(m)costheta=(mgcos(lamdas))/(3m)` <br/> `v(dv)/(dx)=(g)/(3)<a href="https://interviewquestions.tuteehub.com/tag/cos-935872" style="font-weight:bold;" target="_blank" title="Click to know more about COS">COS</a>(lamdas)` <br/> `int_(0)^(v)vdv=(<a href="https://interviewquestions.tuteehub.com/tag/8-336412" style="font-weight:bold;" target="_blank" title="Click to know more about 8">8</a>)/(3)int_(0)^(s)cos(lamdas)ds` <br/> `|(v^(2))/(2)|_(0)^(v)=(g)/(3)(|sin(lamdas)|_(0)^(s))/(lamda)` <br/> `v^(2)=(2g)/(3lamda)sin(lamdas)` <br/> `v=sqrt(2gsintheta)/(3lamda)` <br/> `y-`direction`:` `N+Fsintheta=mg` <br/> `N=mg-(mg)/(3)sintheta` <br/> `(sintheta)_(max)=1,N_(<a href="https://interviewquestions.tuteehub.com/tag/min-548008" style="font-weight:bold;" target="_blank" title="Click to know more about MIN">MIN</a>)=(2mg)/(3)` <br/> Since normal reaction will <a href="https://interviewquestions.tuteehub.com/tag/never-570518" style="font-weight:bold;" target="_blank" title="Click to know more about NEVER">NEVER</a> will never be zero and hence the block will never <a href="https://interviewquestions.tuteehub.com/tag/lose-537625" style="font-weight:bold;" target="_blank" title="Click to know more about LOSE">LOSE</a> constact with the ground.</body></html> | |
| 38093. |
A ball strikes an identical ball with velocity 50 m/s and after the collision which is perfectly elastic it is found to move with velocity 30 m/s . Show that the two balls move at right angles to each other after collision . |
| Answer» | |
| 38094. |
A bigger unit is contained………………….. Of times in the quantity. |
| Answer» <html><body><p><br/></p><a href="https://interviewquestions.tuteehub.com/tag/answer-15557" style="font-weight:bold;" target="_blank" title="Click to know more about ANSWER">ANSWER</a> :<a href="https://interviewquestions.tuteehub.com/tag/smaller-1213327" style="font-weight:bold;" target="_blank" title="Click to know more about SMALLER">SMALLER</a> <a href="https://interviewquestions.tuteehub.com/tag/number-582134" style="font-weight:bold;" target="_blank" title="Click to know more about NUMBER">NUMBER</a></body></html> | |
| 38095. |
A light inextensible smooth string is in contact with a smooth-pulley (fixed at C) through an anglephis shown in figure. If T is the tension in the string, then force with which the string presses the pulley vertically is xTsin ( phi//2)then x is. |
| Answer» <html><body><p><br/></p><a href="https://interviewquestions.tuteehub.com/tag/answer-15557" style="font-weight:bold;" target="_blank" title="Click to know more about ANSWER">ANSWER</a> :<a href="https://interviewquestions.tuteehub.com/tag/2-283658" style="font-weight:bold;" target="_blank" title="Click to know more about 2">2</a></body></html> | |
| 38096. |
Pick out the dimensionless quantity……….. |
| Answer» <html><body><p><a href="https://interviewquestions.tuteehub.com/tag/vapour-1442886" style="font-weight:bold;" target="_blank" title="Click to know more about VAPOUR">VAPOUR</a> density<br/><a href="https://interviewquestions.tuteehub.com/tag/specific-1220917" style="font-weight:bold;" target="_blank" title="Click to know more about SPECIFIC">SPECIFIC</a> <a href="https://interviewquestions.tuteehub.com/tag/gravity-18707" style="font-weight:bold;" target="_blank" title="Click to know more about GRAVITY">GRAVITY</a> <br/>molality<br/>mass fraction</p>Solution :specific gravity</body></html> | |
| 38097. |
An accelration produces a narrow beam of protons, each having an initial speed of v_(0). The beam is directed towards an initially uncharges distant metal sphere of radius R and centered at point O. The initial path of the beam is parallel to the axis of the sphere at a distance of (R//2) from the axis, as indicated in the diagram. The protons in the beam that collide with the sphere will cause it to becomes charged. The subsequentpotential field at the accelerator due to the sphere can be neglected. The angular momentum of a particle is defined in a similar way to the moment of a force. It is defined as the moment of its linear momentum, linear replacing the force. We may assume the angular momentum of a proton about point O to be conserved. Assume the mass of the proton as m_(P) and the charge on it as e. Given that the potential of the sphere increases with time and eventually reaches a constant velue. If the initial kinetic energy of a proton is 2.56 ke V, then the final potential of the sphere is |
| Answer» <html><body><p>`2.56 kV`<br/>`1.92 kV`<br/>greater than `2.56 kV`<br/>needs more information</p>Solution :From <a href="https://interviewquestions.tuteehub.com/tag/previous-592857" style="font-weight:bold;" target="_blank" title="Click to know more about PREVIOUS">PREVIOUS</a> we can see that the final potential <a href="https://interviewquestions.tuteehub.com/tag/energy-15288" style="font-weight:bold;" target="_blank" title="Click to know more about ENERGY">ENERGY</a> of the <a href="https://interviewquestions.tuteehub.com/tag/sphere-1222094" style="font-weight:bold;" target="_blank" title="Click to know more about SPHERE">SPHERE</a> is equal to the `3//4` of initial <a href="https://interviewquestions.tuteehub.com/tag/kinetic-533291" style="font-weight:bold;" target="_blank" title="Click to know more about KINETIC">KINETIC</a> energy <br/> `rArr 3/4xx2.56=1.92 kV`</body></html> | |
| 38098. |
(A) : The velocity of a body at the bottom of an inclined plane of given height, is more when it slides down the plane, compared to when it rolling down the same plane. (R ) : In rolling down a body acquires both kinetic energy of translation and rotation |
| Answer» <html><body><p>Both 'A' and '<a href="https://interviewquestions.tuteehub.com/tag/r-611811" style="font-weight:bold;" target="_blank" title="Click to know more about R">R</a>' and true and 'R' is the correct <a href="https://interviewquestions.tuteehub.com/tag/explantation-2628194" style="font-weight:bold;" target="_blank" title="Click to know more about EXPLANTATION">EXPLANTATION</a> of 'A'<br/>Both 'A' and 'R' and true and 'R' is <a href="https://interviewquestions.tuteehub.com/tag/notthe-2877399" style="font-weight:bold;" target="_blank" title="Click to know more about NOTTHE">NOTTHE</a> correct explantation of 'A'<br/>A' is true and 'R' is <a href="https://interviewquestions.tuteehub.com/tag/false-459184" style="font-weight:bold;" target="_blank" title="Click to know more about FALSE">FALSE</a><br/>A' is false and 'R' is true</p>Answer :A</body></html> | |
| 38099. |
An accelration produces a narrow beam of protons, each having an initial speed of v_(0). The beam is directed towards an initially uncharges distant metal sphere of radius R and centered at point O. The initial path of the beam is parallel to the axis of the sphere at a distance of (R//2) from the axis, as indicated in the diagram. The protons in the beam that collide with the sphere will cause it to becomes charged. The subsequentpotential field at the accelerator due to the sphere can be neglected. The angular momentum of a particle is defined in a similar way to the moment of a force. It is defined as the moment of its linear momentum, linear replacing the force. We may assume the angular momentum of a proton about point O to be conserved. Assume the mass of the proton as m_(P) and the charge on it as e. Given that the potential of the sphere increases with time and eventually reaches a constant velue. The limiting electric potential of the sphere is |
| Answer» <html><body><p>`(3m_(P)v_(0)^(<a href="https://interviewquestions.tuteehub.com/tag/2-283658" style="font-weight:bold;" target="_blank" title="Click to know more about 2">2</a>))/(8e)`<br/>`(3m_(P)v_(0)^(2))/(4e)`<br/>`(3m_(P)v_(0)^(2))/(2e)`<br/>None of these</p>Solution :Limiting <a href="https://interviewquestions.tuteehub.com/tag/electric-967871" style="font-weight:bold;" target="_blank" title="Click to know more about ELECTRIC">ELECTRIC</a> potential= charge in `DeltaKE` <br/> `(KE)_(i)=1/2 mv_(0)^(2): (KE)_(<a href="https://interviewquestions.tuteehub.com/tag/f-455800" style="font-weight:bold;" target="_blank" title="Click to know more about F">F</a>)=1/2 m(v_(0)/2)^(2)` <br/> `:.` Electric potential `=(3mv_(0)^(2))/(8e)`</body></html> | |
| 38100. |
If the radius of the earth becomes half of its present value but its density remains unchanged then find the weight of body on the surface of earth. |
| Answer» <html><body><p></p>Solution :`implies g =(GM_(e))/R_e^2=G/(R_e^2)(4/3piR_e^3rho)=4/3 piR_erho`<br/>Since the <a href="https://interviewquestions.tuteehub.com/tag/radius-1176229" style="font-weight:bold;" target="_blank" title="Click to know more about RADIUS">RADIUS</a> becomes <a href="https://interviewquestions.tuteehub.com/tag/half-1014510" style="font-weight:bold;" target="_blank" title="Click to know more about HALF">HALF</a>, <br/> `g. = 4/3 piG((R_e)/2)rho =1/2 (4/3 pi GR_erho)` <br/> `:. g. =g/2 ` soweight becomes half <br/> Weight `W <a href="https://interviewquestions.tuteehub.com/tag/prop-607409" style="font-weight:bold;" target="_blank" title="Click to know more about PROP">PROP</a> g` <br/> `:. (W.)/W = (g.)/g` <br/> `:. W. = W <a href="https://interviewquestions.tuteehub.com/tag/xx-747671" style="font-weight:bold;" target="_blank" title="Click to know more about XX">XX</a> 1/2` <br/> ` :. W. = W/2`</body></html> | |