ChemistryQ51–54 of 84 questions
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51
ChemistryMCQAIPMT · 2004
Which one of the following pairs is not correctly matched?
A
$Streptomyces$ - Antibiotic
B
$Serratia$ - Drug Addiction
C
$Spirulina$ - Single cell protein
D
$Rhizobium$ - Biofertilizer
Solution
(B) $Streptomyces$ is a genus of bacteria known for producing various antibiotics.
$Spirulina$ is widely used as a source of single-cell protein $(SCP)$ due to its high nutritional value.
$Rhizobium$ is a well-known nitrogen-fixing bacterium used as a biofertilizer.
$Serratia$ is a genus of Gram-negative, facultatively anaerobic bacteria. Some species, like $Serratia$ $marcescens$, are opportunistic human pathogens known to cause urinary tract infections, wound infections, and pneumonia. It is not associated with drug addiction.
$Spirulina$ is widely used as a source of single-cell protein $(SCP)$ due to its high nutritional value.
$Rhizobium$ is a well-known nitrogen-fixing bacterium used as a biofertilizer.
$Serratia$ is a genus of Gram-negative, facultatively anaerobic bacteria. Some species, like $Serratia$ $marcescens$, are opportunistic human pathogens known to cause urinary tract infections, wound infections, and pneumonia. It is not associated with drug addiction.
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52
ChemistryMCQAIPMT · 2004
In India,electricity is supplied for domestic use at $220 \,V$. It is supplied at $110 \,V$ in the $USA$. If the resistance of a $60 \,W$ bulb for use in India is $R$,what will be the resistance of a $60 \,W$ bulb for use in the $USA$?
A
$R$
B
$2 R$
C
$\frac{R}{4}$
D
$\frac{R}{2}$
Solution
(C) The power rating of a bulb is given by the formula $P = \frac{V^2}{R}$,where $V$ is the voltage and $R$ is the resistance.
For the bulb used in India: $R = \frac{V_{India}^2}{P} = \frac{(220)^2}{60}$.
For the bulb used in the $USA$: $R' = \frac{V_{USA}^2}{P} = \frac{(110)^2}{60}$.
Dividing the two equations: $\frac{R}{R'} = \frac{(220)^2}{(110)^2} = \left(\frac{220}{110}\right)^2 = 2^2 = 4$.
Therefore,$R' = \frac{R}{4}$.
For the bulb used in India: $R = \frac{V_{India}^2}{P} = \frac{(220)^2}{60}$.
For the bulb used in the $USA$: $R' = \frac{V_{USA}^2}{P} = \frac{(110)^2}{60}$.
Dividing the two equations: $\frac{R}{R'} = \frac{(220)^2}{(110)^2} = \left(\frac{220}{110}\right)^2 = 2^2 = 4$.
Therefore,$R' = \frac{R}{4}$.
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53
ChemistryMCQAIPMT · 2004
An animal phylum having radially symmetrical adults but bilateral symmetrical larvae is
A
Porifera
B
Coelenterata
C
Echinodermata
D
Annelida
Solution
(C) The phylum $Echinodermata$ is characterized by a unique feature where the adult forms exhibit radial symmetry (specifically pentamerous radial symmetry),while their larval stages exhibit bilateral symmetry.
This transition from bilateral symmetry in larvae to radial symmetry in adults is a diagnostic feature of this phylum.
Therefore,the correct option is $C$.
This transition from bilateral symmetry in larvae to radial symmetry in adults is a diagnostic feature of this phylum.
Therefore,the correct option is $C$.
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54
ChemistryMCQAIPMT · 2004
If $|\vec{A} \times \vec{B}| = \sqrt{3} \vec{A} \cdot \vec{B}$,then the value of $|\vec{A} + \vec{B}|$ is
A
$\left(A^2 + B^2 + \frac{AB}{\sqrt{3}}\right)^{1/2}$
B
$A + B$
C
$\left(A^2 + B^2 + \sqrt{3}AB\right)^{1/2}$
D
$\left(A^2 + B^2 + AB\right)^{1/2}$
Solution
(D) Given: $|\vec{A} \times \vec{B}| = \sqrt{3} \vec{A} \cdot \vec{B}$
Using the definitions of cross and dot products: $AB \sin \theta = \sqrt{3} AB \cos \theta$
Dividing both sides by $AB \cos \theta$: $\tan \theta = \sqrt{3}$
Therefore,the angle between the vectors is $\theta = 60^{\circ}$
The magnitude of the resultant vector is given by: $|\vec{A} + \vec{B}| = \sqrt{A^2 + B^2 + 2AB \cos \theta}$
Substituting $\theta = 60^{\circ}$: $|\vec{A} + \vec{B}| = \sqrt{A^2 + B^2 + 2AB \cos 60^{\circ}}$
Since $\cos 60^{\circ} = 1/2$: $|\vec{A} + \vec{B}| = \sqrt{A^2 + B^2 + 2AB(1/2)} = \sqrt{A^2 + B^2 + AB}$
Thus,$|\vec{A} + \vec{B}| = (A^2 + B^2 + AB)^{1/2}$
Using the definitions of cross and dot products: $AB \sin \theta = \sqrt{3} AB \cos \theta$
Dividing both sides by $AB \cos \theta$: $\tan \theta = \sqrt{3}$
Therefore,the angle between the vectors is $\theta = 60^{\circ}$
The magnitude of the resultant vector is given by: $|\vec{A} + \vec{B}| = \sqrt{A^2 + B^2 + 2AB \cos \theta}$
Substituting $\theta = 60^{\circ}$: $|\vec{A} + \vec{B}| = \sqrt{A^2 + B^2 + 2AB \cos 60^{\circ}}$
Since $\cos 60^{\circ} = 1/2$: $|\vec{A} + \vec{B}| = \sqrt{A^2 + B^2 + 2AB(1/2)} = \sqrt{A^2 + B^2 + AB}$
Thus,$|\vec{A} + \vec{B}| = (A^2 + B^2 + AB)^{1/2}$
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