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Question

If the straight lines x=1+s,y=-3-λs,z=1+λs and x equals fraction numerator t over denominator 2 end fraction comma y equals 1 plus t, z=2-t, with parameters s and t respectively, are co planar, then λ=

  1. -2    
  2. 0    
  3. fraction numerator negative 1 over denominator 2 end fraction    
  4. -1    

The correct answer is: -2

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If 5 cas 2 theta plus 2 Cos squared space left parenthesis theta divided by 2 right parenthesis plus 1 equals 0 comma 0 less than theta less than pi not stretchy rightwards double arrow theta equals

If 5 cas 2 theta plus 2 Cos squared space left parenthesis theta divided by 2 right parenthesis plus 1 equals 0 comma 0 less than theta less than pi not stretchy rightwards double arrow theta equals

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A boy begins to walk eastward along a street infront of his house and the graph of his displacement from home is shown in the following figure. His average speed for in the whole-time interval is equal to

D i s p l a c e m e n t blank f o r m blank 0 blank t o blank 5 blank s blank equals blank 40 blank m
D i s p l a c e m e n t blank f r o m blank 5 blank t o blank 10 blank s blank equals blank 40 blank m
D i s p l a c e m e n t blank f r o m blank 0 blank t o blank 15 blank s blank equals blank minus blank 20 blank m
A n d blank d i s p l a c e m e n t blank f r o m blank 15 blank t o blank 20 blank s equals 0 blank m
therefore N e t blank d i s p l a c e m e n t equals 40 plus 40 minus 20 plus 0 equals 60 blank m
T o t a l blank t i m e blank t a k e n equals 5 plus 5 plus 15 plus 5 equals 30 blank m i n.
H e n c e comma blank a v e r a g e blank s p e e d equals fraction numerator d i s p l a c e m e n t blank open parentheses m close parentheses over denominator t i m e blank open parentheses m i n close parentheses end fraction equals fraction numerator 60 over denominator 30 end fraction equals 2 m blank m i n to the power of negative 1 end exponent.

A boy begins to walk eastward along a street infront of his house and the graph of his displacement from home is shown in the following figure. His average speed for in the whole-time interval is equal to

physics-General
D i s p l a c e m e n t blank f o r m blank 0 blank t o blank 5 blank s blank equals blank 40 blank m
D i s p l a c e m e n t blank f r o m blank 5 blank t o blank 10 blank s blank equals blank 40 blank m
D i s p l a c e m e n t blank f r o m blank 0 blank t o blank 15 blank s blank equals blank minus blank 20 blank m
A n d blank d i s p l a c e m e n t blank f r o m blank 15 blank t o blank 20 blank s equals 0 blank m
therefore N e t blank d i s p l a c e m e n t equals 40 plus 40 minus 20 plus 0 equals 60 blank m
T o t a l blank t i m e blank t a k e n equals 5 plus 5 plus 15 plus 5 equals 30 blank m i n.
H e n c e comma blank a v e r a g e blank s p e e d equals fraction numerator d i s p l a c e m e n t blank open parentheses m close parentheses over denominator t i m e blank open parentheses m i n close parentheses end fraction equals fraction numerator 60 over denominator 30 end fraction equals 2 m blank m i n to the power of negative 1 end exponent.
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The values of theta satisfying Sin space 5 theta equals Sin space 3 theta minus Sin space theta and 0 less than theta less than pi over 2 are

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A body of mass m is resting on a wedge of angle theta as shown in figure. The wedge is given at acceleration alpha. What is the value of a son that the mass m just falls freely?

The horizontal acceleration a of the wedge should be such that in time the wedge moves the horizontal distance B C. The body must fall through a vertical distance A B under gravity. Hence,
B C equals fraction numerator 1 over denominator 2 end fraction a t to the power of 2 end exponent and A B equals fraction numerator 1 over denominator 2 end fraction g t to the power of 2 end exponent
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A body of mass m is resting on a wedge of angle theta as shown in figure. The wedge is given at acceleration alpha. What is the value of a son that the mass m just falls freely?

physics-General
The horizontal acceleration a of the wedge should be such that in time the wedge moves the horizontal distance B C. The body must fall through a vertical distance A B under gravity. Hence,
B C equals fraction numerator 1 over denominator 2 end fraction a t to the power of 2 end exponent and A B equals fraction numerator 1 over denominator 2 end fraction g t to the power of 2 end exponent
tantheta equals fraction numerator A B over denominator B C end fraction equals fraction numerator g over denominator a end fractionor a equals fraction numerator g over denominator t a n theta end fraction equals g blank c o t theta
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A cyclist starts from the centre O of a circular park of radius one kilometre, reaches the edge P of the park, then cycles along the circumference and returns to the centre along Q O as shown in figure. If the round trip takes ten minutes, the net displacement and average speed of the cyclist (in metre and kilometre per hour) is

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equals 1 plus fraction numerator 2 pi cross times 1 over denominator 4 end fraction plus 1 equals fraction numerator 14.28 over denominator 4 end fraction blank k m
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A cyclist starts from the centre O of a circular park of radius one kilometre, reaches the edge P of the park, then cycles along the circumference and returns to the centre along Q O as shown in figure. If the round trip takes ten minutes, the net displacement and average speed of the cyclist (in metre and kilometre per hour) is

physics-General
Net displacement equals 0 and total distance equals O P plus P Q plus Q O
equals 1 plus fraction numerator 2 pi cross times 1 over denominator 4 end fraction plus 1 equals fraction numerator 14.28 over denominator 4 end fraction blank k m
Average speedblank equals fraction numerator 14.28 over denominator 4 cross times 10 divided by 60 end fraction
equals fraction numerator 6 cross times 14.28 over denominator 4 end fraction equals 21.42 blank k m divided by h
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physics-General
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I is not possible because total distance covered by a particle increases with time
II is not possible because at a particular time, position cannot have two values
III is not possible because at a particular time, velocity cannot have two values
IV is not possible because speed can never be negative

Which of the following graphs cannot possibly represent one dimensional motion of a particle

physics-General
I is not possible because total distance covered by a particle increases with time
II is not possible because at a particular time, position cannot have two values
III is not possible because at a particular time, velocity cannot have two values
IV is not possible because speed can never be negative
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Which of the following graphs cannot possibly represent one dimensional motion of a particle

I is not possible because total distance covered by a particle increases with time
II is not possible because at a particular time, position cannot have two values
III is not possible because at a particular time, velocity cannot have two values
IV is not possible because speed can never be negative

Which of the following graphs cannot possibly represent one dimensional motion of a particle

physics-General
I is not possible because total distance covered by a particle increases with time
II is not possible because at a particular time, position cannot have two values
III is not possible because at a particular time, velocity cannot have two values
IV is not possible because speed can never be negative
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Solution of Tan space x plus Tan space open parentheses 120 to the power of 6 plus x close parentheses plus Tan space open parentheses x minus 120 to the power of ring operator close parentheses equals 0 is

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If Tan space bold A plus Tan space 2 bold A plus square root of 3 Tan space bold ATan 2 bold A equals square root of 3 then the general solution of A over 2 equals

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