PhysicsQ1–2 of 2 questions
Page 1 of 1 · English
1
PhysicsEasyMCQAIIMS · 1996
Three vectors $\vec a$,$\vec b$,and $\vec c$ satisfy the relations $\vec a \cdot \vec b = 0$ and $\vec a \cdot \vec c = 0$. The vector $\vec a$ is parallel to:
A
$\vec b$
B
$\vec c$
C
$\vec b \cdot \vec c$
D
$\vec b \times \vec c$
Solution
(D) Given that $\vec a \cdot \vec b = 0$,this implies that the vector $\vec a$ is perpendicular to the vector $\vec b$.
Given that $\vec a \cdot \vec c = 0$,this implies that the vector $\vec a$ is perpendicular to the vector $\vec c$.
By the definition of the cross product,the vector $\vec b \times \vec c$ results in a vector that is perpendicular to both $\vec b$ and $\vec c$.
Since $\vec a$ is perpendicular to both $\vec b$ and $\vec c$,and $\vec b \times \vec c$ is also perpendicular to both $\vec b$ and $\vec c$,it follows that $\vec a$ must be parallel to $\vec b \times \vec c$.
Given that $\vec a \cdot \vec c = 0$,this implies that the vector $\vec a$ is perpendicular to the vector $\vec c$.
By the definition of the cross product,the vector $\vec b \times \vec c$ results in a vector that is perpendicular to both $\vec b$ and $\vec c$.
Since $\vec a$ is perpendicular to both $\vec b$ and $\vec c$,and $\vec b \times \vec c$ is also perpendicular to both $\vec b$ and $\vec c$,it follows that $\vec a$ must be parallel to $\vec b \times \vec c$.
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2
PhysicsMediumMCQAIIMS · 1996
If $_{92}U^{238}$ undergoes successively $8 \alpha$-decays and $6 \beta$-decays,then the resulting nucleus is
A
$_{82}U^{206}$
B
$_{82}Pb^{206}$
C
$_{82}U^{210}$
D
$_{82}U^{214}$
Solution
(B) The initial nucleus is $_{92}U^{238}$.
An $\alpha$-decay reduces the mass number $A$ by $4$ and the atomic number $Z$ by $2$.
$A$ $\beta$-decay does not change the mass number $A$ and increases the atomic number $Z$ by $1$.
After $8 \alpha$-decays,the new mass number $A' = 238 - (8 \times 4) = 238 - 32 = 206$.
The new atomic number $Z' = 92 - (8 \times 2) + (6 \times 1) = 92 - 16 + 6 = 82$.
The element with atomic number $82$ is Lead $(Pb)$.
Therefore,the resulting nucleus is $_{82}Pb^{206}$.
An $\alpha$-decay reduces the mass number $A$ by $4$ and the atomic number $Z$ by $2$.
$A$ $\beta$-decay does not change the mass number $A$ and increases the atomic number $Z$ by $1$.
After $8 \alpha$-decays,the new mass number $A' = 238 - (8 \times 4) = 238 - 32 = 206$.
The new atomic number $Z' = 92 - (8 \times 2) + (6 \times 1) = 92 - 16 + 6 = 82$.
The element with atomic number $82$ is Lead $(Pb)$.
Therefore,the resulting nucleus is $_{82}Pb^{206}$.
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