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    Chemistry-
    General
    Easy

    Question

    Which of the following covalent molecule is an exception to octet rule?

    1. BeCl2    
    2. CO2    
    3. H2O    
    4. CH4    

    The correct answer is: BeCl2

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    Valence of sulphur in sulphuric acid is

    Valence of sulphur in sulphuric acid is

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    The maximum valence of an element with atomic number 7 is

    The maximum valence of an element with atomic number 7 is

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    Metal ‘M’ forms a peroxide of the type MO2 . Valency of the metal with respect to oxygen

    Metal ‘M’ forms a peroxide of the type MO2 . Valency of the metal with respect to oxygen

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    A heavy mass is attached to a thin wire and is whirled in a vertical circle. The wire is most likely to break

    Because here tension is maximum

    A heavy mass is attached to a thin wire and is whirled in a vertical circle. The wire is most likely to break

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    Because here tension is maximum
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    Consider a vector stack F with rightwards arrow on top equals 4 stack i with hat on top minus 3 stack j with hat on top. Another vector that is perpendicular to stack F with rightwards arrow on top is

    Since stack F with rightwards arrow on top equals 4 stack i with hat on top minus 3 stack j with hat on top is lying in X minus Y plane, hence the vector perpendicular to stack F with rightwards arrow on top must be lying perpendicular toX minus Y plane i e comma along Z-axis.

    Consider a vector stack F with rightwards arrow on top equals 4 stack i with hat on top minus 3 stack j with hat on top. Another vector that is perpendicular to stack F with rightwards arrow on top is

    physics-General
    Since stack F with rightwards arrow on top equals 4 stack i with hat on top minus 3 stack j with hat on top is lying in X minus Y plane, hence the vector perpendicular to stack F with rightwards arrow on top must be lying perpendicular toX minus Y plane i e comma along Z-axis.
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    A circular road of radius 1000 blank m has banking angle 45 degree. The maximum safe speed of a car having mass 2000 blank k g will be, if the coefficient of friction between tyre and road is 0.5

    The maximum velocity for a banked road with friction,
    v to the power of 2 end exponent equals g r open parentheses fraction numerator u plus t a n theta over denominator 1 minus u t a n theta end fraction close parentheses
    rightwards double arrow v to the power of 2 end exponent equals 9.8 cross times 1000 cross times open parentheses fraction numerator 0.5 plus 1 over denominator 1 minus 0.5 cross times 1 end fraction close parentheses rightwards double arrow v equals 172 m divided by s

    A circular road of radius 1000 blank m has banking angle 45 degree. The maximum safe speed of a car having mass 2000 blank k g will be, if the coefficient of friction between tyre and road is 0.5

    physics-General
    The maximum velocity for a banked road with friction,
    v to the power of 2 end exponent equals g r open parentheses fraction numerator u plus t a n theta over denominator 1 minus u t a n theta end fraction close parentheses
    rightwards double arrow v to the power of 2 end exponent equals 9.8 cross times 1000 cross times open parentheses fraction numerator 0.5 plus 1 over denominator 1 minus 0.5 cross times 1 end fraction close parentheses rightwards double arrow v equals 172 m divided by s
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    NH3 has a much higher boiling point than PH3 because

    NH3 has a much higher boiling point than PH3 because

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    The maximum number of molecules that one water molecule can hold through hydrogen bonding is

    The maximum number of molecules that one water molecule can hold through hydrogen bonding is

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    The wavelength of the wave function of a bonding molecular orbital formed by LCAO is

    The wavelength of the wave function of a bonding molecular orbital formed by LCAO is

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    In the formation of homo diatomic neutral molecule, if ‘N’ atomic orbitals combine, then the total number of bonding molecular orbitals formed is

    In the formation of homo diatomic neutral molecule, if ‘N’ atomic orbitals combine, then the total number of bonding molecular orbitals formed is

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    An object is moving in a circle of radius 100 m with a constant speed of 31.4 blank m s to the power of negative 1 end exponent. What is its average speed for one complete revolution?

    The time taken by the particle for one complete revolution.
    t equals fraction numerator 2 pi r over denominator s p e e d end fraction
    equals fraction numerator 2 cross times 3.14 cross times 100 over denominator 31.4 end fraction equals 20 s
    H e n c e comma blank a v e r g e blank s p e e d blank i s blank
    v subscript a v end subscript equals fraction numerator 2 cross times 3.14 cross times 100 over denominator 20 end fraction equals 31.4 blank m s to the power of negative 1 end exponent

    An object is moving in a circle of radius 100 m with a constant speed of 31.4 blank m s to the power of negative 1 end exponent. What is its average speed for one complete revolution?

    physics-General
    The time taken by the particle for one complete revolution.
    t equals fraction numerator 2 pi r over denominator s p e e d end fraction
    equals fraction numerator 2 cross times 3.14 cross times 100 over denominator 31.4 end fraction equals 20 s
    H e n c e comma blank a v e r g e blank s p e e d blank i s blank
    v subscript a v end subscript equals fraction numerator 2 cross times 3.14 cross times 100 over denominator 20 end fraction equals 31.4 blank m s to the power of negative 1 end exponent
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    The horizontal range of a projectile 4 square root of 3 times the maximum height achieved by it, then the angle of projection is

    fraction numerator u to the power of 2 end exponent sin invisible function application 2 theta over denominator g end fraction equals 4 square root of 3 cross times fraction numerator u to the power of 2 end exponent sin invisible function application theta over denominator 2 g end fraction
    or fraction numerator u to the power of 2 end exponent over denominator g end fraction 2 sin invisible function application theta cos invisible function application theta equals 2 square root of 3 fraction numerator u to the power of 2 end exponent over denominator g end fraction sin to the power of 2 end exponent invisible function application theta

    The horizontal range of a projectile 4 square root of 3 times the maximum height achieved by it, then the angle of projection is

    physics-General
    fraction numerator u to the power of 2 end exponent sin invisible function application 2 theta over denominator g end fraction equals 4 square root of 3 cross times fraction numerator u to the power of 2 end exponent sin invisible function application theta over denominator 2 g end fraction
    or fraction numerator u to the power of 2 end exponent over denominator g end fraction 2 sin invisible function application theta cos invisible function application theta equals 2 square root of 3 fraction numerator u to the power of 2 end exponent over denominator g end fraction sin to the power of 2 end exponent invisible function application theta
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    physics-

    Two wires A C and B C are tied at C of small sphere of mass 5 kg, which revolves at a constant speed v in the horizontal circle of radius 1.6 m. The minimum value of v is

    From force diagram shown in figure
    T subscript 1 end subscript cos invisible function application 30 degree plus T subscript 2 end subscript cos invisible function application 45 degree equals m g (i)
    T subscript 1 end subscript sin invisible function application 30 degree plus T subscript 2 end subscript sin invisible function application 45 degree equals fraction numerator m v to the power of 2 end exponent over denominator r end fraction (ii)
    After solving Eq. (i) and eq. (ii), we get
    T subscript 1 end subscript equals fraction numerator m g minus fraction numerator m v to the power of 2 end exponent over denominator r end fraction over denominator open parentheses fraction numerator square root of 3 minus 1 over denominator 2 end fraction close parentheses end fraction
    But T subscript 1 end subscript greater than 0
    therefore fraction numerator m g minus fraction numerator m v to the power of 2 end exponent over denominator r end fraction over denominator fraction numerator square root of 3 minus 1 over denominator 2 end fraction end fraction greater than 0

    or m g greater than fraction numerator m v to the power of 2 end exponent over denominator r end fraction
    or v less than square root of r g end root
    therefore v subscript m a x end subscript equals square root of r g end root equals square root of 1.6 cross times 9.8 end root equals 3.96 blank m s to the power of negative 1 end exponent

    Two wires A C and B C are tied at C of small sphere of mass 5 kg, which revolves at a constant speed v in the horizontal circle of radius 1.6 m. The minimum value of v is

    physics-General
    From force diagram shown in figure
    T subscript 1 end subscript cos invisible function application 30 degree plus T subscript 2 end subscript cos invisible function application 45 degree equals m g (i)
    T subscript 1 end subscript sin invisible function application 30 degree plus T subscript 2 end subscript sin invisible function application 45 degree equals fraction numerator m v to the power of 2 end exponent over denominator r end fraction (ii)
    After solving Eq. (i) and eq. (ii), we get
    T subscript 1 end subscript equals fraction numerator m g minus fraction numerator m v to the power of 2 end exponent over denominator r end fraction over denominator open parentheses fraction numerator square root of 3 minus 1 over denominator 2 end fraction close parentheses end fraction
    But T subscript 1 end subscript greater than 0
    therefore fraction numerator m g minus fraction numerator m v to the power of 2 end exponent over denominator r end fraction over denominator fraction numerator square root of 3 minus 1 over denominator 2 end fraction end fraction greater than 0

    or m g greater than fraction numerator m v to the power of 2 end exponent over denominator r end fraction
    or v less than square root of r g end root
    therefore v subscript m a x end subscript equals square root of r g end root equals square root of 1.6 cross times 9.8 end root equals 3.96 blank m s to the power of negative 1 end exponent
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    Two stones are projected with the same velocity in magnitude but making different angles with the horizontal. Their ranges are equal. If the angel of projection of one is pi divided by 3 and its maximum height is y subscript 1 end subscript, the maximum height of the other will be

    Given theta subscript 1 end subscript equals pi divided by 3 equals 30 degree
    Horizontal range is same if theta subscript 1 end subscript plus theta subscript 2 end subscript equals 90 degree
    therefore theta subscript 2 end subscript equals 90 degree minus 30 degree equals 60 degree
    y subscript 1 end subscript equals fraction numerator u to the power of 2 end exponent sin to the power of 2 end exponent invisible function application 30 degree over denominator 2 g end fraction and y subscript 2 end subscript fraction numerator u to the power of 2 end exponent sin to the power of 2 end exponent invisible function application 60 degree over denominator 2 g end fraction
    therefore fraction numerator y subscript 2 end subscript over denominator y subscript 1 end subscript end fraction equals fraction numerator sin to the power of 2 end exponent invisible function application 30 degree over denominator sin to the power of 2 end exponent invisible function application 60 degree end fraction equals open parentheses fraction numerator 1 divided by 4 over denominator square root of 3 divided by 4 end fraction close parentheses to the power of 2 end exponent blank equals fraction numerator 1 over denominator 2 end fraction blank o r blank y subscript 2 end subscript equals fraction numerator y subscript 1 end subscript over denominator 3 end fraction

    Two stones are projected with the same velocity in magnitude but making different angles with the horizontal. Their ranges are equal. If the angel of projection of one is pi divided by 3 and its maximum height is y subscript 1 end subscript, the maximum height of the other will be

    physics-General
    Given theta subscript 1 end subscript equals pi divided by 3 equals 30 degree
    Horizontal range is same if theta subscript 1 end subscript plus theta subscript 2 end subscript equals 90 degree
    therefore theta subscript 2 end subscript equals 90 degree minus 30 degree equals 60 degree
    y subscript 1 end subscript equals fraction numerator u to the power of 2 end exponent sin to the power of 2 end exponent invisible function application 30 degree over denominator 2 g end fraction and y subscript 2 end subscript fraction numerator u to the power of 2 end exponent sin to the power of 2 end exponent invisible function application 60 degree over denominator 2 g end fraction
    therefore fraction numerator y subscript 2 end subscript over denominator y subscript 1 end subscript end fraction equals fraction numerator sin to the power of 2 end exponent invisible function application 30 degree over denominator sin to the power of 2 end exponent invisible function application 60 degree end fraction equals open parentheses fraction numerator 1 divided by 4 over denominator square root of 3 divided by 4 end fraction close parentheses to the power of 2 end exponent blank equals fraction numerator 1 over denominator 2 end fraction blank o r blank y subscript 2 end subscript equals fraction numerator y subscript 1 end subscript over denominator 3 end fraction
    General
    physics

    Angle of projection, maximum height and time to reach the maximum height of a particle are theta,H and  respectivley. Find the true relation.

    Angle of projection, maximum height and time to reach the maximum height of a particle are theta,H and  respectivley. Find the true relation.

    physicsGeneral
    parallel

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