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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.
| 12451. |
A mass of 4 kg rests on a horizontal plane. The plane is gradually inclined until at an angle 0 = 15^(@) with the horizontal, the mass just begins to slide. What is the coefficient of static friction between the block and the surface |
Answer» SOLUTION : Forceactingon blockand theircomponentare shown in figure When `theta `increasestan `theta`will alsoincreasehencemagnitudeof `f_(s) ` will increaseuptocertainlimitLetfor `theta = theta_(max)` frictionalforcebecomemaximum`f_(s) _(max)` fromequatin(3) `tan theta _(max)` `tan 15^(@) =mu_(s)` `mu_(s)= tan15^(@)= 0.2679` `mu_(s)= 0.27` |
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| 12452. |
Four particles each of mass 100 g are placed at the corners of a square of side 10 cm. Find the moment of inertia of the system about an axis passing through the centre of the square and perpendicular to its plane. Find also the radius of gyration of the system |
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Answer» Solution :mass `=100 g` radius `r=5 sqrt(2)cm` Momentum of inertia `1I=4 mr^(2)` `=4xx100xx(5 sqrt(2))^(2)=2XX10^(-3) kgm^(2)` `K=sqrt((I)/(M))= sqrt((2xx10^(-3))/(0.4))= 0.07m =7 cm` |
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| 12453. |
A fixed cylindrical vessel is filled with water up to height H. A hole is bored in the wall at a depth h from the free surface of water. For maximum horizontal range .h. is equal to H/N where N is equal to 2. |
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| 12454. |
A wooden rod of a uniform cross section and of length 120 cm in hinged at the bottom of the tank which is filled with water to a height of 40cm. In the equilibrium position, the rod makes an angle of 60^(@) with the vertical. The centre of buoyancy is located on the rod at a distance (from the hinge) of |
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Answer» 20 cm |
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| 12455. |
A girl riding a bicycle along a straight road with a speed of 5 ms^(-1) throws a stone of mass 0.5 kg which has a speed of 15 ms^(-1) with respect to the ground along her direction of motion.The mass of the girl and bicycle is 50 kg. Does the speed of the bicycle change after the stone is thrown ? What is the change in speed, if so |
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Answer» Solution :Massof girlbicycle =50 kg TOTALMASS of system(girl+bicycle+ stone ) VELOCITYOF bicycle `mu_(1)m//s`mass of stone`m_(2) =0.5 kg` Velocityof stone`u_(2) =15m//s ` Massof girlandbicyclem= 50kg Yes the speedof the bicyclewill changeafter thestone is thrown. Let afterthrowingthe stone`vm//s`speedof bicycle By lawof conservationof linearmomentum `m_(1)u_(1) = m_(2) u_(2)+mv` 50.5 XX 5 =0.5xx 15+ 50xx v` 252.5=7.5= 50 v` `v= (245.0)/( 50)` `v= 4.9 m//s` Changein speed=5-4.9=0.1m//s` |
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| 12456. |
The focal lengths of objective lens and eyelens of a Gallelian Telescope are respectively 30cm and 3.0cm. Telescope produces virtual,erect image of an object situated far away from it at least distance an object situated far away from it at least distance of distinct vision from the eye lens. In this condition , the Magnifying power of the Gallelian Telescope should be: |
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Answer» `+11.2` |
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| 12457. |
A satellite is revolving in a stable orbit of radius r with orbital velocity v_(0) around the earth. The time period of the satellite is T, angular momentum L and its total energy is E.Then which one of the following statement is not correct? |
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Answer» `V_(0)` is DIRECTLY proportional to `R^((-3)/(2))` |
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| 12458. |
Find the anglebetween vec(P) = - 2hat(i) +3 hat(j) +hat(k) and vec(Q) = hat(i) +2hat(j) - 4hat(k) |
| Answer» SOLUTION :`THETA = 90^(@)` | |
| 12459. |
An equilateral triangle ABC is formed by two copper rods AB and BC and one is aluminium rod which heated in such a way that temperature of each rod increases by DeltaT. Find change in the angle angleABC. (Coefficient of linear expansion for copper is alpha _(1). and for aluminium is alpha2.). |
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Answer» Solution :Suppose,`AB = l _(1), AC =_(2) and BC = l _(3)` `therefore COS theta = ( l _(3) ^(2) + l _(1) ^(2) - l _(2) ^(2))/( 2 l _(3) l _(1))`where ` angleABC = theta ` `therefore 2l _(3) l _(1) cos theta = l _(3) ^(2) + l_(1) ^(2) - l _(2) ^(2)`  Integrating on both side, `2 ( l _(3) dl _(1) + l _(1) dl _(3)) cos theta- 2 l3l _(1) sin theta d theta =2l _(3) dl _(3) +2l _(1) dl _(1) -2l _(2) dl _(2)` Dividing by 2, `(l_(3) dl _(1) + l _(1) xx dl _(3)) cos theta -l _(3) l _(1) sin theta d theta =l_(3) dl _(3) + l _(1) dl _(1) - l _(2) dl _(2)` Now `dl _(1) = l _(1) alpha _(1) Delta T, dl _(2) = l _(1) alpha _(2) Delta T, dl _(3) = l _(3) alpha _(3) Delta T ` then, `(l _(3) xx l _(1) alpha _(1) Delta T + l _(1) xx l _(3) alpha _(3) Delta T) cos theta - l _(3) l _(1) sin theta d theta =` `l _(3) xx l _(3) alpha _(3) Delta T + l_(1) xx l _(1) alpha _(1) Delta T - l _(2) xx l _(2) alpha _(2) Delta T` Now let `l _(1) =l _(2) = l _(3) = l and alpha _(3) = alpha _(1)` `therefore (l ^(2) alpha _(1) Delta T + l ^(2) alpha _(1) Delta T) cos theta - l ^(2) sin theta d theta = l ^(2) alpha _(1) Delta T +^(2) alpha _(1) Delta T - l ^(2) alpha _(2) Delta T` `cos theta = cos 60^(@) = 1/2 ` (Equilateral triangle) `therefore 2l ^(2) alpha _(1) Delta T xx 1/2 - l ^(2) sin theta d theta = 2l alpha _(1) Delta T - l ^(2) alpha _(2) Delta T` `therefore l alpha _(1) Delta T - l ^(2) sin theta d theta = 2l ^(2) alpha _(1) Delta T - l ^(2) alpha _(2) Delta T ` Dividing by `l ^(2),` `alpha _(1) Delta T - sin thetad theta = 2 alpha _(1) Delta T - alpha _(2) Delta T` `- alpha _(1) Delta T + alpha _(2) Delta T = sin theta d theta ` `therefore d theta = ((alpha _(2) - alpha _(1))Delta T )/( sin theta )` `therefore d theta =(2 (alpha _(2) - alpha _(1)) Delta T)/( SQRT3) [ because sin theta = sin 60^(@) = (sqrt3)/(2) ]` |
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| 12460. |
We can shield a change from electric fields by a putting it inside a hollow conductor.Can we shild a body from the gravitational influence of nearby matter by putting it inside a hollow sphere or by some other means? |
| Answer» Solution :No, because gravitational force between two point mass bodies is independent of the interventing medium between them. It is due the above REASON, we cannot shiled a body from the gravitational influence of NEARBY MATTER by putting it either INSIDE a hollow sphere or by some other means. | |
| 12461. |
A satellite is revolving round the earth with orbital speed v_(0). IF its stops suddenly, the speed with which it will strike the surface of earth would be sqrt(v_(e )^(2)-kv_(0)^(2)) (v_(e ) - escape velocity of a particle on earth.s surface ). FInd the value of K |
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| 12462. |
The total random kinetic enrgy of 1g of helium at 100 K will be (R=8.3"Jmol1.^(-1)K^(-1)) |
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Answer» 622.50J |
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| 12463. |
Speeds of two identical cars are u and 4u at a specific instant. The ratio of the respective distances in which two cars are stopped from that instant. |
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Answer» `1:1` `THEREFORE = ((0)^(2) - u ^(2))/( 2A)` `therefore d =- ( u^(2))/( 2a),` here `-(1)/(2a) ` is CONSTANT. `therefore d alpha u ^(2)` `therefore (d _(1))/(d _(2)) = ((u _(1))/( u _(2)))^(2) = ((u)/( 4u )) ^(2)` `therefore (d _(1))/(d_(2)) = (1)/(16)` `therefore d _(1) : d _(2) =1 :16` |
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| 12464. |
A particle is executing SHM with amplitude A. At what distance from mean position will the KE and PE are equal ? |
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Answer» SOLUTION :SINCE KE = PE, `1/2 mv^2 =1/2 m omega^2 x^2 implies1/2 m omega^2 (A^2 - x^2) =1/2 m omega^2 x^2` `impliesA^2 - x^2 = x^2 ` (or) `x= A/(sqrt2)` |
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| 12465. |
A vertical hollow cylinder of height 1.52 m is fitted with a movable piston of negligible mass and thickness. The lower half portion of the cylinder contains an ideal gas and the upper half is filled with mercury. The cylinder is initially at 300 K. When the temperature is raised, half of the mercury comes out of the cylinder. Find this temperature, assuming the thermal expansion of mercury to be negligible. |
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Answer» Solution :Initially, Pressure = 76 + 76 = 152 cm of HG From PV = nRT `(152xx76xxA)/300=nR""...(1)` FINAL pressure = 76 + 38 = 114 cm of Hg `therefore(114xx(76+38)A)/T=nR""...(2)` From (1) and (2) `(152xx76)/300=(114xx114)/T` `T=(300xx114xx114)/(152xx76)=(300xx3)/4xx3/2=337.5K` 
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| 12466. |
Apparent weight of a piece of metal immersed in water at 0^(@)C is 100 g and at 50^(@)C is 100.5 g. What will be the apparent weight at 20^(@)C? |
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Answer» 100.1 g |
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| 12467. |
Find the tensions in the string (1),(2) and (3) and the acceleration of the mass 'm' just after ( initially system is in equilibrium and at rest, pulley, string, spring are light ): (a)string (1) is cut (b) string (2) is cut (c ) string (3) is cut |
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Answer» `(B)` `T_(1)=0,T_(2)=T_(3)=0,a=g` `(c )` `T_(1)=0,T_(2)=0,T_(3)=0,a=g` |
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| 12468. |
Find the fundamental, first overtone and second overtone frequencies of an open organ pipe of length 20 cm. Speed of sound in air is 340 m s^-1. |
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Answer» `I=20cmxx10^-2m` Fundamental frequency `=(V/21)=340/(2xx20xx10^-2)` `=850 Hz` We known first over tone `((2V)/(2I))=(2xx340)/(2xx10xx10^-2) ` (for open pipe)` `=1750Hz` SECOND OVERTONE `=3(V/2I)=3xx850=2500Hz` |
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| 12469. |
When body is at rest or it is in uniform motion, no force act on it. |
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Answer» SOLUTION :Whebnbodyisatrestor inuniformmotiongravitationalforceacton it. Forceslikefrictionforceviscousforceact ONBODY. Whenobjectis at restinuniformmotionon it iszero. l Considerbookat RESTON horizontalsurfacefollowingforceact on its. Herebookis atresthenceW=R Actuallyit SHOULDBE likethisconsideringbookat restbyNewtonfirstlawof motionresultantof externalforceactingon bodymustbe zerothissuggestthatnormalforceand weightshouldbeequaland inoppositedirection 
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| 12470. |
A satellite of mass m revolving in a circular orbit of radius 3 R_Earound the earth (mass of earth is Me and radius is R_E ). How much excess energy be spent to bring it to orbit of radius 9 R_E ? |
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Answer» `(GM_Em)/(3R_E)` `:.` Energy at `r = 3R_E` `|E_1|=(GM_Em)/(2(3R_E))=(GM_Em)/(6R_E)` and Energy at `r = 9R_E` `|E_2|=(GM_Em)/(2(9R_E))=(GM_Em)/(18R_E)` `:.` EXCESS energy from going `3R_E " to " 9R_E` `DeltaE=E_1 -E_2` `= (GM_Em)/(6R_E) - (GM_Em)/(18R_E)` `= (2GM_Em)/(18R_E)` `= (GM_Em)/(9R_E)` |
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| 12471. |
The figure represent a longitudianl waave lengthtravellingin positivex-direction. Then |
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Answer» PART ABC REPRESENT combination |
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| 12472. |
A spring is cut into 4 equal parts & 2 parts are connected in parallel. What is the effective in parallel. What is the effective spring constant. |
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Answer» 4 K |
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| 12473. |
You move forward when your car suddenly comes to a halt and you are thrown backward when your car rapidly accelerates. Which law of Newtons is involved in these ? |
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Answer» THIRD law |
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| 12474. |
What are the angles of contact for wetting and non wetting liquids? |
| Answer» SOLUTION :LESS than `90^(@)`, more than `90^(@)`. | |
| 12475. |
The radius of a sphere is measured as (10+- 0.02%)cm. The error in the measurement of its volume is |
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Answer» 25.1cc |
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| 12476. |
If vec(a) = hat(i) + hat(j) + 2 hat(k) and vec(b) = 3 hat(i) + 2 hat(j) - hat(k), find the value of(vec(a) + 3 vec(b)). ( 2 vec(a) - vec(b)). |
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Answer» Solution :(a) T - MG = ma (B) `T_(x) +F_(x) = A_(x) +B_(x)` (C)`T_(x) - F_(x) = A_(x) - B_(x)` (d)T + mg = ma |
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| 12477. |
The kinetic energy of a particle moving along a straight line is propotioanl to the time .t. of its travel. Then its acceleration is proprtional to |
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Answer» `(1)/SQRT(t)` |
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| 12478. |
A particle is projected up with initial speed u=10 m s^(-1) from the top of a bitlding at time t=0. At time t=5 s the particle strikes the fround. Find the height of the building in meter. . |
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Answer» Solution :First the particle moes up and reaches to highest point . Then the particle moves downward direction and finally strikes ground aftre `5 s`. At highest point the speed of the particle will be zero. Hence, speed is in upward direction and acceleration is in dounward direction. Hence, the sign of acceleration should be negative. USING `v^(2)=u^(2)+2as` Hence, `0=(10)^(2)-2xx10xxH_(1) rArrH_(1)=5 m`  Using `v =u+` at from `A` to `B`, `0=10-10xxt_(1) rArr t_(1)=1 s` At higherst starts, the particle will be zero. The particle starts moving towards downward directon. Hene the velocity any acceleration both are in downward direction. Hence, the sing of acceleration should be pownward . The time taken by the particle from ` B` to `C` will be `5-1=4 s` Now using `s=ut+(1)/(2)^(2)` from `B` to `C` `(H_(1)+H)=(1)/(2)xx10xx4^(2)=80 m` `5+H=80 m` `rArr` Henceheight of the bjjuilding `H=75 m` Using vector method The particle starts from `A` and finally reaches at `C `. Let us take the origin at `A`. The upward direction is takenas positive and the downward dirction is toaen as netative.  . The particle moves in gravitational field where acceleration due to gravity is ALWAYS ACTING at downward direction whether it is movingupward or downward. Hence, acceleration vector `vec a` will always be `-10 m s^(2)`, as its marnitude as well as direction remain constant always THROUGHOUT the motion. Hence, acceleration `vec a =-10 m s^(-2)` Initially at `t=0`, the oartucke is prouected in upward direction. Hence, initial vellocity `vec u=10 m s^(-1)` The particle moves from `A` to `B` (upward) and then `B` to `C` (upwnward). The motion of the particle from `A` and `B` then again passes point `A`. The net displacement of the particle upto this instant is zero. Then particle crosses Point `A` and finally reaches to `C`. We know net displacement is wqual to the difference of final position vector and initial postition vector. Hence, net displacemecnt of the particle during motion `t=5 s` is `-H m`. Using `vec s = vec ut +(1)/(2) vec at^(2)` `=(10)xx5+(1)/(2)(-10)(5)^(2)=50-125=-75 m` Hence, `H=75 m` . |
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| 12479. |
Along a streamline ……. |
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Answer» the velocity of a fluid particle remains constant.  In STREAMLINE flow , velocity of each particle at each point remains constant The velocity of particle PASSING through point is `vecv_(A)`and its DIRECTION is the direction of tangent drawn at that point .It will be for all points. |
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| 12481. |
A particle projected horizontally from P, so that it hits the inclined plane perpendicularly Time of journey is |
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Answer» `SQRT((2H COS^(2)theta)/(g(1+SIN^(2)theta)))` |
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| 12482. |
A geo-stationary satellite is orbiting the Earth of a height of 6R above the surface ofEarth R being the radius of the Earth calculate the time period of another satellite at a height of 2.5R from the surface of Earth. |
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Answer» SOLUTION :Distanceof SATELLITE fromthe centreare 7R and 3.5 R respectively . `(T_(2))/(T_(1)) = ((a_(2))/(a_(1)))^(3/2)` `T_(2) = T_(1) ((a_(2))/(a_(1)))^(3/2) = 24 ((3.5R)/(7R) )^(3/2)` `T_(2) = 8.49 ` (or) `6sqrt(2)` hours |
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| 12483. |
A uniform solid sphere is released from the top of a fixed inclined plane of inclination 30^(@) and height h. It rolls without sliding. The acceleration of the centre of the sphere is |
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Answer» `(3G)/(5)` |
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| 12484. |
The temperature of the sun measured with |
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Answer» PLATINUM thermometer |
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| 12485. |
A uniform solid sphere is released from the top of a fixed inclined plane of inclination 30^(@) and height h. It rolls without sliding. The time taken by the sphere to reach the bottom is |
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Answer» `sqrt((2H)/(G))` |
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| 12486. |
The dimensional formula of the physical quantity whose S.I unit is Hm^(-1) is |
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Answer» `ML^(2) T^(-2) I^(-2)` |
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| 12487. |
The maximum velocity of the particle executing S.H.M is .V. If the amplitude is doubled and the time period of oscillation decreased to 1/3 of its original value, the maximum velocity becomes |
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Answer» 18 V |
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| 12488. |
If the Earth's satellite is put into an orbit at a height where resistance due to atmosphere, can not be neglected, how will motion of satellite be affected? |
| Answer» Solution :Air resistance will reduce the ORBITAL SPEED of satellite. Hence, the satellite will stars decribing spiral path of DECREASING RADIUS and ulimetely falls back to the Earth. Air resistance may also produce a lot of HEAT and the satellite may aven burn. | |
| 12489. |
The internal energy of a system changes whenit undergoes |
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Answer» a cyclic PROCESS |
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| 12490. |
The elongation of a spring of length 'L' and of negligible mass due to a force is 'x'. The spring is cut into two pieces of length in ratio 1: n. The ratio of the respective spring constants is |
| Answer» ANSWER :A | |
| 12491. |
Two masses of 3 and 4 kg are connected at the two ends of a light inextensible that passes over a frictionless pulley. Find the acceleration of the masses and the tension in the string, when the masses are released. |
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Answer» SOLUTION :(i) `a=((M_(1)-M_(2))g)/(M_(1)+M_(2))=((4-3)9.8)/(4+3)=1.4 ms^(-1)` (ii) `T=(2M_(1)M_(2)g)/(M_(1)+M_(2))=(2(4)(3)(9.8))/(4+3)=33.6 N`. |
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| 12492. |
A source whistling a sound of frequency 450 Hz moves towards a stationary listener with a speed of 33 m/s. If velocity of sound is 333 m/s, then find frequency of sound heard by this listener. |
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Answer» 409  We have, `(f _(L))/( v + v _(L)) = (f _(S))/( v + v _(S))` `THEREFORE ( f _(L))/( 333 + 0) = (450)/(333-33)` `therefore f _(L) = 450 XX (333)/(300)= 499.5Hz ~~ 500HZ` |
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| 12493. |
In the above problem if the system can stay in equilibrium. So, the value of F is "x" xx 3N . Where 'x' is |
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| 12494. |
A heavy sphere of mass m is suspended by a string of length l. The spheres is made to revolve about a verticle line passing through the point of suspension, in a horzinontal circle such that the string alwys remins inclined to the verticle making an angle theta What is the period of revolution ? |
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Answer» `T=2pisqrt(L/(9))` |
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| 12495. |
A ship of length l-150m moving with velocity v_(s)=36 km h^(-1) on the sea, suddenly discovered straight head a siking boat people having met an acceleident. A rescue boat has been lowered from the mid of the ship, which went to the sinking boat with speed v_(b)=72 km h^(-1). When the rescue boat was x_(0)=3.0 km away, The rescue boat reaches the sinking boat spends t_(0)=1.0mim there to take the people on board, and then retuned with the same speed to the time taken in the whole rescue it was lowerd. Derermine the time taken in the whole rescue operation from the moment the rescue boat was lowerd to the moment therescue boat returned to the ship. |
Answer» Solution :Time taken by rescue boat to teach from `C` to`E`  . `t_(10=(75))/(v_(B)-v_(s)) =(75)/(20-10) =7.5s` Time taken by rescue boat to GO from `E to `S`: `t_(2) =(300)/(v_(B)) =(3000)/(20) =150 s` Time spent `t_(0)=1 min =60 s Distance travelled by ship time, `t_(2)+t_(0)=v_(s)(15+60) =10 xx 210 =2100 m` Time taken by rescue boat to REACH from `S` to `C`, `t_(3)=((3000-2100)+75)/(v_(B)+v_(s)) =(975)/(30) =32.5 s` Total time : `T=t_(1)+t_(2)+t_(0)+t_(3)`. |
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| 12496. |
Solid cylinders of radii r_(1),r_(2) and r_(3) roll down an inclined plane from the same place simultaneously. r_(1)gtr_(2)gtr_(3), which one would reach the bottom first |
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Answer» CYLINDER of radius `r_(1)` |
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| 12497. |
Draw a plots of mechanical energy, potential energyand kineticversus dispalcementfor different position of a motion of blockattached to a spring . |
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Answer» Solution :The EQUATIONOF mechanicalenergyfor a blockattachedto the end of spring at the pointbetween maximum distance`x_(m)` and zero is , `E = 1/2 kx_(m)^(2)` Equationof potential energy is `V(x) = 1/2 kx_(m)^(2)` and equation of KINETICENERGY is `K= 1/2 mv_(m)^(2)` At equilibrium positionx0the speed ismaximumand hencethe kineticenergy ia maximumthat means `1/2 mv_(m)^(2) = 1/2 kx_(m)^(2)`  ` :. ` Graphof energies `to ` displacement 
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| 12498. |
A body is moving unidirectionally under the influence of a source of constant power supplying energy . Which of the diagram shown infigure correctly shownthe displacement time curve for its motion ? |
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Answer»
P = `(dW)/(DT)` ` = (F.DX)/(dt) ""( :. dW = FDX)` ` = ("Fdx")/(dt)` = constant ` ( :. " P = constant given")` Thus ,P = `FV ""( :. V = (dx)/(dt))` [F] [V] = constant ` :. [MLT^(-1)[ [LT^(-1)] ` = constant ` :. L^(2) T^(-3) = ` constant`( :. "mass is constant") ` ` :. L prop T^(3/2) rArr"Displacement " (d) prop t^(3/2)` |
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| 12499. |
The above question if weighing machine reads correct reading then find acceleration of manl. |
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Answer» ` 5m//s^(2)` |
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| 12500. |
In nature, the failure of structural members usually result from large torque becuae of twisting or bending rather thendueto tensile or compressive strains. This process of structural breakdown is called bucking and in case of tell cylindrical structures like tress, the torque is caused by its own weight bending the structure. Thus the vertical through the centre of gravity does not fall within the base. The elastic torque caused because of this bending about the centralaxis of the tree is given by (Y pi r^4)/(4R). Y is the young's modulus, r is the radius of the trunk and R is the radius of curvature of the bent surface along the height of the tree containing the centre of gravity (the neutral surface). Estimate the critical height of a tree for a given radius of the trunk. |
Answer» Solution :The bending torque on the trunk of radius R of tree `= (Y PI r^4)/(4R)` Where R is the radius of curvature of the bent surface. LET h be the height of tree. If R gtgt h, then the centre of gravity of tree is at a height h/2 from the groung.  in `Delta ABC` Since d ltlt R, therefore the term `d^2` being very very small can be NEGLECTED. `:. R^2 = R^2 - 2 Rd + h^2 //4 or d = (h^2)/(8R) ..... (i)` If `W_0` is the weight per UNIT volume of the tree, then `(T pi r^4)/(4R) = W_0 (pi r^2 h)d = W_0 pi r^2 h xx((h^2)/(8R))` from (i) `:' h= ((2Y)/(W_0))^(1//3) r^(2//3)` |
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